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THE ELEMENTS 



■OF- 



Natural Philosophy 



BY- 



J. W. MOORE, A. M., M. D., 



Professor of Mechanics and Experimental Philosophy, 
Lafayette College, 



EASTON, PENNA. 



itiJtl n 



G. W. West, Book and Job Printer, 
i 891. 



Copyright, 1891, by J. W. Moore. 



A 



CONTENTS. 



STATICS. 
A. COMPOSITION OF FORCES. 

I.— THE PARALLELOGRAM OF FORCES, (a) Forces Acting in one 
Plane, i. Forces Acting at a Point, i. Two Inclined Forces Act- 
ing at a Point, §17, 18 ; 2. Any Number of Forces in the Same Line, 
$20,21,22; 3. Any Number of Inclined Forces at a Point, $23 ; 4. Any 
Number by Rectangular Co-ordinates, $69. 11. Forces Not Acting at 
a Point. [1. Parallel.] 1. Two Parallel Forces Acting at Two Points 
in a Rigid Body, §24 ; 2. Any Number of Parallel Forces in One Plane, 
$25, 26, 70. Couples. 1. Couples with the Same Axis, §31, 32, 33, 33a. 

1. Two Couples, $34, 35,36 ; 3. Any Number of Couples, $37. 2. Coup- 
les with Different Axes. 1. Two Couples, $38, 39, 40 ; 2. Any Num- 
ber of Couples, £41 ; 3. Three Couples in Different Planes, $42. [2. In- 
clined.] 1. Force and Couple, $68 ; 2. Any Number of Inclined For- 
ces; 1. Graphical Method, $71; 2 By Rectangular Co-ordinates, $72,73. 

(b) Forces Not in the Same pAlne. i Acting at a Point. 1. Three 
Forces, $43 ; 2. Any Number of Forces, $44, 75. 11. Forces Not Act- 
ing at a Point. [1. Parallel Forces.] 1. Any Number, $45, 76. [2. 
Inclined Forces.] 1. Any Number, $46, 78. 

II —THE PRINCIPLE OF MOMENTS, (a) Forces in a Plane, i. For- 
ces Acting at a Point. 1. Two Inclined, $51, 52, 52a ; 2. Any Num- 
ber Inclined, $53; 3. Any Number on Same Line, $56. 11. Forces 
Not Acting at a Point. [1. Parallel.] 1. Two, $54. 2. Any Number, 
$55- 3- Couples. (2. Inclined). 1. Any Number, $53. 

(b) Forces Not in the Same Plane, i. Forces Acting at a Point. 
i. Three; 2. Any Number. 11. Forces Not Acting at a Point. 1. Par- 
allel ; 2. Inclined. 

B. RESOLUTION. 

I.— BY THE PARALLELOGRAM OF FORCES, (a) Components in 
one Plane, i. At a Point. 1. Force into two Components, $58; 

2. Into any Number of Components, $59 ; 3. Into two Rectangular Com- 
ponents, |6o ; 4. Components along any line, $63. 11. Not at a Point. 

1. Into two or more Parallel Forces, $61. 

(b) Components Not in the Same Plane, i. At a Point. 1. Com- 
ponents along three Rectangular Axes, $62. 2. Not at a Point. 

2. Components are Parallel Forces, $61. 

II.— BY COUPLES. 1. Into a Single Force and a Couple, $64 ; 2. Into a 
Single Force and Two Couples, $65 ; 3. Into a Single Force and Three 
Couples, $66. 



C. BALANCE Of EQUILIBRIUM. 

I.— BY THE PARALLELOGRAM OF FORCES, (a) Forces Acting 
in A Peane. i. Forces Acting at a Point, i. A Single Force, g8i ; 
2. Two in Same Line, §82 ; 3. Any Number in Same Line, $83 ; 4. Two 
Inclined at a Point, $84 ; 5. Three Inclined at a Point, §85, 86 ; 6. Any 
Number Inclined at a Point, $87, 88. 11. Forces Not Acting at a Point. 
1. Any Number Inclined, §89, 90, 91. in. Parallel. 1. Two, ^92; 
2. Three, $93; 3, Any Number, $94. Couples. 1. With Same Axis. 
1. Two, $95; 2. Any Number, $9653 Force, &c , §97. 2. Different 
Axes. 1. Two, $98; 2. Three, $99 ; 3. Any Number, $100. 

(b) Forces Not in Same Peane. i. Forces at a Point. Three, $102 ; 
Any Number, $103. 11. Not at a Point. 1. Parallel; 1. Any Num- 
ber, $104; 2. Any Number whatever, $105. 

Illustrations of the Principle of the Parallelogram of For- 
ces and the Principle of Moments. 

A. — Centre oe Mass. i. Of a Uniform Straight Line, $ 109 ; 2. Of the Per- 
imeter of a Polygon, $110; 3. Of a Parallelogram, $111; 4. Of a Tri- 
angle, $112; 5. Of a Regular Polygon, 1 113 ; 6. Of two Heavy Bodies, 
#114; 7. Of any Number of Heavy Bodies, $115. 115^; 8. Of a Trian- 
gular Pyramid, #116; 9. Of any Pyramid with a Plane Rectilinear 
Polygon as a Base, $117 ; 10. Of a Cone, $118. Properties of the Cen- 
tre of Gravity, $120, 121, 122, 123, 123^7, 124. Equilibrium — Stable, 
Unstable, Neutral, $125, 126, 127. Stability, $128. 

B— Simpee Machines. i. The Lever, £130, 131, 132, 133, 134, 135, 136, 
*37, ! 38, J 39 ; 2. The Wheel and Axle, #140 ; 3. The Rope Machine, 
$141 ; 4. The Pulley, $142 ; 5. The Inclined Plane, §143 ; 6 The Screw, 
§144, 145 ; 7. The Isosceles Wedge, #146. 

C. — Compound Machines, i. The Compound Lever, $148 ; 2. The Knee 
Joint, 1 149, 150; 3. Roberval's Balance, #151 ; 4. The Compound Bal- 
ance, $152; 5. The Compound Wheel and Axle, #153; 6 Wheel 
Work, £154 ; 7. The Differential Wheel and Axle J 155, 156 ; 8. Wes- 
ton's Pulley Block, #157 ; 9. Compound Pulley, $158, 159, 160, 161, 
162, 163, 164, 165 ; 10. The Pulley and Inclined Plane, gi66 ; n. The 
Screw and Lever, \ 167 ; 12. The Differential Screw, $168; 13 Hun- 
ter's Screw, §169 ; 14. The Endless Screw, $170. The Advantage or 
Efficiency of a Machine, §171. 



KINEMATICS. 

Distance, §173, 174. 1. Distances of one Point from another, §175, 176. 

2. Several Points, $177. 

Motion, §178. 1. Motion of two Points, $178-182; 2. Three Points, $183; 

3. Any number of Points, $184. 

Velocity, §185. 1. Uniform, §186, 187, 188, 189. 






Acceleration, $i9 r >. i- Uniformly Accelerated Velocity, $192-206 ; 2. Ve- 
locity of a Body Projected Downward in Vacuo, $207 ; 3. Velocity of 
a Body down an Inclined Plane, §209 ; 5. Projectiles in Vacuo, §210, 
211 ; 6. Uniform Velocity in a Circle, $215-220. 

Angular Velocity, $221-228^. 

Angular Acceleration, 229, 230. 

Motions oe Rigid Bodies. i. Translation, §231; 2. Rotation, #232-242 
Illustrations of the Instantaneous Axis, $243 ; Mechanical Generation 
of the Path, $244 ; Parallel Rotations in the Same Plane, $245-249 ; 
Rotations about Inclined Axes, £250. 






KINETICS. 

Motion. First Law of Motion, #258-260 ; Second Law of Motion, $261, 262, 
263, 264 ; Third Law of Motion, $267. 

Applications of the Second Law oe Motions, $265, 266. 1. Gravitation, 
$268, 269. Attraction of Uniform Sphere on Exterior Particle, $270 ; 
Attraction of Shell on Particle on Interior, #271 ; Attraction inside a 
Solid Sphere, $272; Attraction on the Surfaces of Spheres $273. 2 De- 
viating Force, $274-292 ; 3. Simple Harmonic Motion, $293 ; 4. The 
Pendulum, #294-304; 5. The Conical Pendulum, $305; 6. The Theory 
of Atwood's Machine, $306. 



THE SCIENCE OF ENERGY. 

Work, $307-311. Unit of Work, $312,313; Work in Lifting a System of 
Bodies, $314; Work in Terms of Pressure and Volume, $315 ; Work 
in Terms of Angular Velocity, $316 ; Work of a System of Moving 
Bodies, $317-319; Work of Acceleration of Translation, $321; Work 
of Acceleration of Rotation, $321-323; Work of Acceleration of Trans- 
lation of a Machine of Several Pieces, $3-3; Work of Acceleration of 
Rotation of a Machine of Several Pieces, $324 ; Work of a Couple, $325. 

Energy. Potential, $326-331 ; Kinetic Energy. Independent of Direction, 
$333 5 Of a System of Bodies, $334-336 ; Of a Rotating Body, $337 ; 
Due to Rectangular Components, $338; Transformation of Energy, 
$339; Availability of Energy, $341 ; Conservation of Energy, $342-344. 



THE THEORY OF POTENTIAL. 

Field of Force, $345 ; Lines of Force, $346, 347 ; Strength of Field at 
a Point, $348, 349, 350; Potential, $351 ; Mutual Potential Energy, 
$35 2 > 353 ! Absolute Potential, $354 ; Direction of Motion, $355 ; Dif- 



ference of Potential, $356; Datum Line, §357; Consequences of the Def- 
inition of Potential, §358 ; Another Form of Absolute Potential, $359 ; 
Another Form of Difference of Potential, §360, 361 ; Equipoteutial 
Surfaces, $362 ; A System of Equipotential Surfaces, $363 ; Geomet- 
rical Principles, $364, 365 ; Flow of Force, $366, 367, 368 ; LaPlace's 
and Poisson's Equations, $369, 370 ; Tubes of Force, §371 ; Properties 
of Tubes of Force, $372, 373, 



DYNAMICS OF LIQUIDS AND GASES. 

Statics of Liquids and Gases, $374 ; Pascal's Principle, §375 ; Principle 
of the Transmissibility of Fluid Pressure, §376; Principle of Perpendicu- 
larity of Pressures, $377-381 ; Pressure on any surface, in any Liquid 
or Gas at any depth, $382 ; Pressure independent of Shape of Vessel 
or quantity of Liquid, $383. Illustrations of Vertical Downward Pres- 
sure, $384 ; Of Vertical Upward Pressure, $385 ; Of Lateral Pressure, 
§386 ; Of Pressure in all Directions, $387 ; Equilibrium of Liquids, 
$388 ; A Level Surface not necessarily a Plane Surface, $389 ; A Sin- 
gle Liquid in Communicating Vessels, $390 ; Different Liquids in 
Communicating Vessels, $391 ; Different Liquids in the Same Vessel, 
§392 ; Equilibrium of Bodies in Gases and Liquids, $393-397 ; Floating 
Bodies, $398, 399 ; Specific Gravity, $400-404 ; Methods of Obtaining 
the Specific Gravity of Liquids, $405-407 ; Scale Hydrometers, $408 ; 
Pressure of Gases, $409 ; Compressibility of Gases, $410 ; Boyle's Law, 
$411; Law of Charles, $412; Kinetics of Liquids and Gases, $413-415 ; 
Head of Water, $416; Torricelli's Law, $417, 418; Discordance be- 
tween Theory and Practice, $419-421 ; Mouthpieces, $422; Efflux from 
Orifices in the Side of a Vessel, $423-425 ; Flow in Open Channels, 
$426 ; Constant Flow in Uniform Rigid Pipes, $427. 



SYMBOLS 



A, the point of application of a force, velocity, &c. 

A, a solid angle. 

a, acceleration, also the area of a cross-section. 

C, centre of moments. 

c, centre of figure. 

F, a force. 

f, the coefficient of friction. 

g, the centre of gravity of a body or system of bodies. 

g, 32.2, the acceleration due to gravity, 
h, vertical height. 

I, the moment of inertia of a whole body. 

i, the moment of inertia of an element. 

A', the coefficient of contraction. 

L, the arm of a couple when not parallel to the co-ordinate 

axes ; also the length of a pendulum. 
M, the moment of a couple, 
m, the mass of a body. 
O, the origin of co-ordinates, 
p, the intensity of a force. 
R, a resultant force. 
S, a surface, 
s, distance. 
T, time. 

t, a very small portion of time. 
t, absolute temperature. 
V, volume. 
V, potential, 
v, "velocity. 
W, weight. 

w, specific weight, or the weight of unit volume. 
X, Y, Z, components or resultants along Ox. 
x, y, z, co-ordinates of a point. 
x, y, z, distances of the centre of gravity of a body measured 

along Ox, Oy, Oz. 
*, the coefficient of expansion of a body. 



o v^ a (The angles a given force, velocity, &c, makes 

' ^ f with the axes Ox, Oy, Oz. 

/- — Gamma ) ' ■ J ' 

^ — Delta. Specific mass or mass per unit volume. 

r — Zeta. The anglethe axis of a resultant couple makes with Oy 

■Eta. " (t " " u " " " Oz 



— Theta. The angle two lines in a plane or in space make 
with each other. 

/ — Lambda I 

ju — Mu /-The angles a resultant makes with Ox, Ov, Oz. 

o—Nu ) 

£ — Xi. The angle the axis of a resultant couple makes with Ox. 

- — Pi. 3. 141592, the circumference of a circle whose diame- 
ter is unity. 

p — Rho. Radius of curvature ; volume density. 

- — Sigma. Sum of. 

<? — Sigma. Surface density. 

<p — Phi. An angle. 

co — Omega. Angular velocity. 

OTHER SYMBOLS. 

=0= " Will produce the same effect,'' read " equivalent to." 
00 Infinity. 

USE OF SUBSCRIPTS. 

When there are several forces, velocities, &c., they are dis- 
tinguished by writing numerals under, as F,, F 2 , F 3 , a n a„ a sy 
/3 3 , /3 2 , /9 S , v„ v 2 , v„ &c. The final resultant in such cases is 
written without a subscript as R, v, X, Y, Z. 

Capital letters are used for components along the axes ; 
small letters for co-ordinates parallel to the axes ; and a dash 
over a small letter is used to represent a final distance in an 
investigation, e. g., x, y, z, are read u line x", "liney!', 
u line z". 

Any force is represented by F, but its nature and direction 
are represented by subscripts, e. g., F g is the force of gravity, 
F d is a deviating force, F s is a stress, F n is a normal force, Ft 
a tangential force ; similarly a t is tangential acceleration, &c. 



ERRATA. 



p. 3, $7, read "in dynamics" instead of " in mechanics." 

p. 5, §4, insert after sign, " or ward." 

p. 8, top of page insert reference — §69. 

P- ! 3> $35- Cor. Read "if the two couples" also "the moment of the re- 
sultant couple.'" 

p. 19, ninth line from bottom at end, add = R . AC. 

p. 23, §68, eighth line, read " Now apply the couple to the single force." 

p. 30, sixth line from the top of page, read " single" instead of " given." 

P- 34. $86, read " If three inclined forces." 

P- 35. $92, add "each other." 

P 35. $93. fifth line, read " distance' 1 '' instead of " distances." 

P- 39. $108, third line, read " ward" instead of " direction." 

p. 44, $121, sixth line, read "supported by a string or bar to'a fixed point O." 

p. 49, second line from the top of the page, read " = a* with the horizon- 
tal line xAxV 

p. 49, sixth line from the bottom of the page in the denomination the first 
minus sign should be changed to plus. 

p. 55, twelfth line from the top, read 1 : 2 cos. 8, 

p. 56, eighth line from the bottom, read b instead of B. 

p. 60, fifth line from the top, read " as twice the height, &c.'' 

p. 64, §166, exchange "height" and "length." 

p. 72, ninth line from bottom, read v* instead of v. 

p. 85, seventh line from bottom read T/T 2 . 



STATICS. 



Definitions. 

i. — Matter is that which occupies space. 

2. — A body is a limited portion of matter. Bodies differ 
from each other in (a) quality, (b) shape, (c) volume or the 
space they occupy, (d) mass or the quantity of matter in them, 
(e) weight or the force with which they are attracted towards 
the earth. 

3. — A particle or molecule is an immeasurably small body. 

4. — A material line may be regarded as composed of an in- 
definite number of particles placed side by side, or as a body 
limited by particles ; when the particles are uniform and uni- 
formly placed the line is said to be uniform. 

5. — A material surface is composed of an indefinite number 
of material lines placed side by side, or is a body limited by 
lines. 

6. — A material volume is composed of an indefinite number 
of material surfaces superposed, or is a body limited by planes. 

7. — In mechanics all bodies are supposed to be rigid, that 
is, their particles have no motion relatively to each other. 

8. — Lines are measured in terms of the yard, foot or inch, 
or fractional parts thereof. If the metric system is used the 
unit of length is the metre. 

Surfaces are measured in square yards, square feet or square 
inches, or square metres or fractional parts thereof. 

Volumes are measured in cubic yards, cubic feet or cubic 
inches, or cubic metres or fractional parts thereof. 

9. — The specific mass of a body or density is the mass of 
unit volume and may be represented by 6. Hence the mass 
(M) of any body equals the product of its volume (V) by its 
specific mass (6) or M=V#. 

The specific weight of a body is the weight of unit volume, 
and may be represented by w. Hence the weight (W) of any 
body equals the product of its volume (V) by its specific 
weight (w) or W=V?^. 



Newton's experiments showed that the weight of a body is 
proportional to its mass, at a given place, or W=^M in which 
g is a constant. 

It follows that the specific weight (w) of a body is propor- 
tional to its specific mass (6) or w=g6 in which g is a con- 
stant. 

10. — Force is an action exerted upon a body in order to 
change its state, either of rest or of moving uniformly forward 
in a straight line. 

ii. — Dynamics is the science offeree ; it is subdivided into 
Statics which treats of balanced forces or of bodies at rest and 
Kinetics, which shows the relation between motion and the 
force which produces it. 

12. — Kinematics is the science of motion without reference 
to its cause. 

13. — There are two kinds of quantities ; to one kind the idea 
of direction is essential. These are called vector quantities. 
Distance, velocity, acceleration, force, impulse axe vectors. 

To the other kind the idea of direction is not attached, such 
are time, mass, volume, speed, potential. These are called 
scalar quantities. 

Any quantity may be represented by a line, but vector quan- 
tities can only be represented by a line drawn in a particular 
direction. 

Hence a vector may be represented by a line (1) whose 
length is proportional to its magnitude, (2) whose sign is de- 
termined by an arrow head placed at the end, (3) and whose 
direction is the direction of the line determined by the prin- 
ciples of Analytical Geometry. 

14. — A force is known when these four things are given : 

1. The point ox place of applicatio7i. 

In actual cases the force is applied to a space which 
may be a surface as in stress or a volume as in weight. 
The intensity of a force is the amount of the force per 



unit surface or per unit volume or p = — or p =~- 

The position of the point of application is given in 
terms of its co-ordinates according to the principles 
of Analytical Geometry. 

2. The magnitude is measured in terms of some unit, 
the pound, for example, or the kilogramme. 

3. The line of action or the line along which motion 
would take place if the body were free to move. 

The direction of this line is given in terms of the 
angles it makes with assumed axes according to the 
principles of Analytical Geometry. . 

4. The sign or direction in the line of action is the di- 
rection of motion of the body. 

15. — Since & force is a vector quantity it may be represented 
by a line (1) the length of which is proportional to the magni- 
tude of the force, (2) the direction of which is the direction of 
the line of action of the force, and (3) the sign of which is rep- 
resented by an arrow-head attached to the line. 

For example, let B C D (Fig. 1) represent a body acted upon 
by a force Fi ; Ai is its point of application. The direction 
of Ai Fi represents its line of action — the arrow indicates the 
direction in the line of action — and if one of the small divi- 
sions represent the force of a pound the line will represent a 
force of six pounds. 

It is evident that the point of application may be moved 
along the line of action without affecting the magnitude or di- 
rection of the force. This truth is sometimes called the prin- 
ciple of the tr admissibility of force. 

A. COMPOSITION OF FORCES. 

16. — Definitions. — The resultant of any number of forces is 
a single force which will produce the same effect as the given 
forces. 

The given forces are called components. 

Composition is the process of finding the resultant. 



I. THE PARALLELOGRAM OF FORCES. 

(a). Forces Acting in one Plane. 

i. Forces Acting at a Point. 

i. Two Inclined Forces Acting at a Point. 

17. — If the two adjacent sides of a parallelogram represent 
in magnitude and direction, two forces acting upon a particle, 
the diagonal drawn from the intersection of these two sides, 
will represent the resultant in magnitude and direction. This 
proposition is called u The Parallelogram of Forces. " 

The truth of this proposition may be taken as the result of ex- 
periment. Let B and C (Figs. 2 & 3) be two smooth pegs over 
which a cord is passed. At the knot A suspend a weight Ri. At- 
tach similar weights Fi and F 2 to the ends of the cord and al- 
low the system to come to rest. On A C cut off a part pro- 
portional to Fi, similarly on A B a part proportional to F 2 . 
Complete the parallelogram A Fi R F 2 , measure A R and it 
will be found that its length is proportional to Ri. 

18. — Cor. 1. If two forces acting upon a particle be repre- 
sented in magnitude a7id direction by tivo sides of a triangle, 
the third side will represent the resultant in magnitude and 
direction. 

Let A Fx and A F 2 (Fig. 4) represent the two forces acting 
upon A. By the "parallelogram of forces" §17, A R will repre- 
sent their resultant. Since AFiRF 2 is a parallelogram AF 2 = 
FiR. Hence AFi and FiR will represent the two forces and 
AR their resultant. 

19. — For purposes of numerical calculation if the two forces 
are given in magnitude and direction, the resultant may be 
found in magnitude and direction by the rules of trigonome- 
try. The following is convenient : 

R2 = F^+F 2 2 -h2F!F 2 cos FiAF 2 . 

2. Any Number of Forces in the Same Line. 

20. — Cor. 2. When the angle between the two adjacent sides 
of the parallelogram becomes zero, the two sides coincide and 



the diagonal equals their sum ; hence the resultant of two for- 
ces in the same line of action and in the same direction equals 
the sum of the two forces. When the included angle becomes 
180 the diagonal equals the difference of the two sides ; hence 
the resultant of two forces in the same line is equal to their 
difference, if opposite to each other in direction, or generally, 
the resultant of two forces in the same straight line equals 
their algebraic sum. 

21. — It is evident that if the two forces are equal and oppo- 
site their resultant will equal zero, and the two forces will 
balance each other. 

22. — Cor. 3. The truth of Cor. 2 can be extended to any 
number of forces in the same straight line — since the result- 
ant of two may be found, then the resultant of this resultant 
and the next force and so on until all are treated. 

Hence generally the restdtant of any number of forces in the 
same straight line equals their algebraic sum. 

L,et Fi, F 2 , — F 3 , — F 4 (Fig. 5) be forces having the same 
line of action. 

Then Fi+F 2 =Ri ; R— F 3 =R 2 ; R 2 — F 4 =:R. 
. " . R=R 2 — F4— Ri — F3 — F4=Fi-|-F 2 — F3 — F4. 

j. Any Number of Inclined Forces at a Point. 

23. — Cor. 4. If all the forces acting upon a particle be rep- 
resented in mag7titude and direction by all the sides of a poly- 
gon taken i7i 'order, except one, the excepted side will represent 
the resultant of the given forces in magnitude and direction. 

Let Fi, F 2 , F 3 , F 4 (Fig. 6) represent forces acting upon A. Find 
the resultant of F x and F 2 by the "parallelogram of forces," 
it will be Ri. Find the resultant of Ri and F 3 , it will be R 2 . 
Find the resultant of R 2 and F 4 , it will be R. Hence R is the 
resultant of Fi, F 2 , F 3 , F 4 . 

Now, in the parallelogram AFiRiF 2 , FiRi=AF 2 ; similarly 
RiR 2 =AF 3 and R 2 R=AF 4 . Therefore if AFi, F1R1, RiR 2 , R 3 R, 
the sides of a polygon, taken in order, represent the forces, 
the remaining side AR will- represent the resultant. 



8 

For purposes of calculation this method is tedious — a shorter 
method will be given further on. 

2. Forces not Acting at a Point. 

i. Two Parallel Forces acting on two points in a rigid body. 

24. — Cor. 5. The resultant of two parallel forces is [/] par- 
allel to them, [2] equals their algebraic sum, [j] divides the 
distance between them ijiversely as their magnititdes, \f\ and 
its sign is determined by that of their algebraic sum. 

Case I. Let the two parallel forces act in opposite direc- 
tions. 

Let Fi and F 2 (Fig. 7) acting at Ai and A 2 two points in a 
rigid body, be any two forces. Produce their lines of action 
until they meet at A. From AAi and AA 2 produced, cut parts 
AN and AM proportional to the forces ; then AR will be the 
resultant of the two given forces. 

[1]. Now Fi : F 2 : : AN : AM : : sin. MAR : sin. NAR : : 

GA ' G A * ' G& ' GQl - 

[2]. Also R : Fi : : AR : AN : : sin. ANR=-sin. MAN : sin. 

MAR : : --i-? 1 :~ :: A 2 P X : A 2 P. 

A 2 A A 2 A 

Now suppose that F x and F 2 become parallel, a condition 
expressed by making the angle MAN=o° or 180 . Expres- 
sion [2] becomes sin. MAR=o, therefore, MAR=o. That is — 

1. R will be parallel to F 2 and therefore parallel to Fi, that 
is, the resultant will be parallel to the two parallel forces. 

Expression [1] will become — 

2. Fi : F 2 : : GQ 2 : GQi : : GA 2 : GAi that is, the point of 
application of the resultant will divide the distance between 
the components inversely as their magnitudes. 

Expression [2] will become — 

R : Fi : : A 2 Pi : A 2 P : : A 2 Ai : GA 2 , which proves the same 
thing. 

. . R A 2 A X A 2 G— AiG AiG F 2 Fi— F 2 

Again rATG = -AS- =I "A^ =I -R = -F-" 



9 

3. R=Fj — F 2 , that is, the magnitude of the resultant is the 
difference of the magnitudes of the given forces. 

Case II. Let the two parallel forces act in the same direc- 
tion. (Fig. 8.) A similar demonstration will show that the 
magnitude of the resultant equals the sum of the magnitudes 
of the components, hence the truth of the proposition. 

2. Any number of parallel forces in one plane. 

25. — Cor. 1. To find the resultant of any number of paral- 
lel forces, find the resultant of two, then the resultant of this 
resultant and the next force, and so on. 

LetFi, F 2 , F 3 , F 4 , (Fig. 9) with points of application Ai, A 2 , 
A 3 , A 4 , be the given forces. Then by §24 the resultant of V x 
and F 2 is 

R^Fi-K— F 2 ). 

The resultant of Ri and F 3 is 
R 2 =F!+(— F,)+(— F,). 

The resultant of R 2 and F 4 is 

[1]. R=F t -F 2 — F3+F4. 

Hence the magnittide of the resultant equals the algebraic 
sum of the magnitudes of the components. 

[2]. The point of application of the resultant is found by 
continued applications of [3] of the preceding proposition. 

Fx :F 2 ::A 2 Oi:A 1 Oi. 

V x : Fx— F 2 :: A 2 Oi : A 2 Oi— A1O1— AiA,. 

Ri :F 3 :: A3O2 : 2 Oi. 

Ri : Rx+F 3 :: A 3 2 : A 3 2 +0 2 0i=A 3 0i, &c. 

[3]. The direction of the resultant in its line of action or 
its sign is determined by the sign of its magnitude. 

For shorter method see §70. 

26. — Cor. 2. The resultant of tivo equal, opposite, parallel 
forces equals zero, and its point of application is at an infinite 
distance from either force. 

Let Fi and F 2 (Fig. 10) be two such forces, then by §24 

R=Fi — F 2 =o; that is, the magnitude of the resultant is 
zero, or, there is no single resultant. 



IO 

The point of application of the resultant divides the distance 
between the forces inversely as their magnitudes, so that if x 2 
and x 1 are the distances of the lines of action of the forces 
from the point of application of the resultant, 

F x :— F 2 ::x 2 :x v 

F x :Fi— F 2 ::x 2 ixj+x^ 

Fi : o : : x 2 : x 2 +xi. 

Fi(x 2 + X a ) .,,..., • • r - f 

x 2 — =00; that is, the arm is infinity. 

Hence the proposition of §24 fails and some other method 
of treatment of such forces must be found. 

COUPLES. 

27. — Definitions. Two forces which are [1] equal, \2\ par- 
allel^ [3] opposite in direction, and [4] whose lines of action 
are not coincident, form a combination called a couple. 

The tendency of a couple is, obviously, to produce rotation 
of the body, upon which it acts, in the plane containing the 
lines of action of the two forces, which plane is called the 
plane of the couple. 

The direction of rotation is clockwise — right-handed as in 
Fig. 10, or counter-clockwise — left-handed as in Fig. 12. 
Right-handed couples are reckoned positive, left-handed, neg- 
ative. 

The axis of the couple is any line perpendicular to its 
plane. 

The arm of the couple is the perpendicular distance be- 
tween the lines of action of the two forces. 

The moment of a force, called torque, relatively to any 
point in its plane is the product of the force and the perpen- 
dicular distance of its line of action from the point. 

Thus, let F (Fig. n) be the given force and C any point in 
its plane ; from C draw CP perpendicular to the line of action 
of the force ; then F-CP is the required moment. 

C is called the centre of moments. CP is called the arm of 
the force F. 



II 

Since force is measured in pounds, and distance in feet, 
moments are measured in foot-pounds. 

28. — Prop. The moment of a couple in reference to any point 
in its plane is the product of the magnitude of either force and 
its arm. 

Let F 1} F 2 , (Fig. 12) be the two forces of the couple and AiA 2 
its arm ; take d as the centre of moments ; then the total mo- 
ment will be ,— Fi- Aid— F 2 -A 2 C 1 =-F 1 (AiC 1 +A 2 C 1 )=-FrA 1 A 2 . 
Again, take C 2 as the centre of moments, then the total moment 
will be; F 2 ■ AiC— F,- K^=V X (A X C 2 — A 2 C 2 )=— F^AtA,). 

29. — A couple is known when there are given, 

1. The ?nagnitude of its moment. 

2. The position of its axis. 

3. The direction of the couple, that is, whether right- 

handed or left-handed. 

30. — To represent a couple by a line : 

1. Draw a line perpendicular to the plane of the couple, 

2. Cut off a part of the line proportional to its moment, 

and 

3. Look along the line so that the couple shall appear 

right-handed. 
For example, let F,L, (Fig. 13) be a couple. From C draw 
a line CP perpendicular to the plane of the couple ; cut off a 
part Cx proportional to the moment and look from C towards 
P so that the couple may appear right-handed. Then Cx rep- 
resents the couple in every respect. 

/. Couples with the same axis, or couples in the same 
. or parallel planes. 

31. — Prop. I. A cotple may be turned round the end of its 
arm , in its own plane, without changing its effect. 

Let F l5 AjA 2 (Fig.- 14) be a couple ; rotate it around A 2 to 
the new position A-jA^ ; the effect of the couple is not changed. 
For at A 2 apply two equal and opposite forces, F 3 and F 4 , each 
equal to V, ; and at A',, F 5 and F (; each equal to F,. The ap- 



12 

plication of these forces will not change the statical condition 
of the body, since their resultant equals zero. 

Find the resultant R x of F 2 and F 4 , also the resultant R 2 of 
F x and F 5 . R 2 and R x are equal and opposite ; their resultant 
equals zero, and F 3 at A 2 and F 6 at A/ remain. 

32. — Prop. II. The arm of a couple may be moved parallel to 
itself in the plane of the couple without changing the effect of 
the couple. 

Let F x , A a A 2 (Fig. 15) be a couple ; move its arm to A^A^ 
parallel to itself in the plane of the couple ; its effect remains 
unchanged, for at A\ and A 2 r apply two equal and opposite 
forces, each equal and parallel to F 3 . Complete the parallelo- 
gram A l A 1 / A / 2 A 2 and draw its diagonals intersecting at O. R, 
is the resultant of F 2 at A 2 and F 4 at A\. R 2 is the resultant 
of F, at A, and F 5 at A 2 r . But the resultant of R, and R 2 equals 
nothing and there remain F 3 at A', and F 6 at A' 9 . 

33. — Prop. III. Any couple may be replaced by another whose 
moment is the same, provided (/) the planes are the same, or 
parallel, (2) the arms are coincident, (3) the extremity of the 
arm common to the two couples, (4) the direction of the couples 
the same. 

Let F 1? AjA 2 (Fig.' 16) be a couple; it may be replaced by 
another couple F 3 • A^g^F, • A, A 2 . 

At A 3 apply two equal and opposite forces F 3 , F 4 , and at A,, 
F 5 and F 6 , all equal to each other. 

Now, the resultant of F 2 and F 4 equals— F 2 -f-F 4 at A,. — F 2 +F 4 
at A l combined with F« equals F 2 . But this is equal and op- 
posite to F x ; hence, there remain F 5 at A, and F 3 at A 3 , a 
couple whose moment F 3 ■ A 1 A 3 =F 1 . A a A 2 . 

33a. — Therefore, to find a couple equivalent to a given couple, 
F 2 L 2 , but with a force Fi, let x be the unknown arm of the equiv- 

F I 
alent couple; then it is obvious, that Fj • x=F L.„ and x=- 2 ,- 2 ' 

F, 
that is, the arm of the equivalent couple is equal to the mo- 
ment of the given couple divided by the given force. 

The foregoing propositions make it possible to find the re- 



sultant of any number of couples in the same or parallel 
planes. 

34. — Prop. IV. The resultant of two couples in the same {or 
parallel) planes is a couple whose moment equals the algebraic 
sum of the moments of the two component couples. 

Let F>L X (Fig. 17) and F 2 L 2 (Fig. 18) be the two couples. 

By proposition I, F 2 L 2 may be turned in position as in (Fig. 

19). 

By Prop. III. it may be made equivalent to a couple (Fig. 

F L 
20) of force Fj but of arm x== — ~ !.. 

Fi 

Apply the new couple to F^ as in (Fig. 21). F, and — F, 

at A 3 have no resultant and F x at A, and — F, at A 2 with arm 

F L 
L r + x == Iy x -f- -^ remain. The moment of this couple equals 

35. — Cor. If the couples have equal moments and are oppo- 
site in direction the moments of the resultant couples will be 
zero. 

$6. — Sometimes it is convenient to find an equivalent cou- 
ple of equal arm instead of equal force. 

Let F^ (Fig. 22) and F 2 L 2 (Fig. 23) be the two couples. 
Reduce F 2 L 2 to an equivalent couple of equal arm (Lj as fol- 
lows : 

F I 
F 2 L 2 = L l F x . F x = -^- 2 that is, the force of the reduced 

equivalent couple equals the moment of the couple divi- 
ded by the given arm. 

The foregoing proposition can then be proved as follows : 

By Prop. II. F 2 L 2 may be moved so that its arm shall co- 
incide with Li, as in (Fig. 240). 

F L 

Now, adding, F x + F X =F X -f-- ^= 2 = resultant force at A l 

and A 2 , with the arm L r 



i4 
The moment-- (p^?^) L, = F I L 1 + F 2 L 2 . 

3J . — Prop. V. The resultant of any number of couples in 
the same or parallel planes is a couple whose moment equals 
the algebraic sum of the moments of the component couples. 

i. Reduce the couples to equivalent couples of equal arm 
or force. 2. Turn them into position with arms and extrem- 
ities of arms common, or forces common. 3. Add the forces, 
or arms. 4. Multiply the sum of the forces or arms by the 
common arm or force. 

Or, find the resultant of two couples, then the resultant of 
this resultant and the next couple by continued applications of 
Prop. IV, §34, as follows : 

M -F^+F^ ; M 2 =M 1 +F 3 Ls; M 3 =M 2 +F 4 L 4 ; 
M=M r fF 3 L3+F 4 L4=F 1 L 1 +F 2 L 2 +FaL3+F4L 4 . 

2. Couples with different axes. 

1. Two couples. 

38. — Prop. If the two adjacent sides of a parallelogram 
represent the moments, positions of axes, and directions of two 
couples acting upon a body, the diagonal draivn from the inter- 
section of these two sides will represent the resultant couple in 
moment, position of axis and direction. This proposition is 
called the "Parallelogram of Couples." 

Let A, B (Fig. 25) be two couples, the intersection of whose 
planes is perpendicular to the plane xyz. Reduce the couples 
to equivalent couples having the common force Fi= F 2 . Let 
a b— Li be the arm of one reduced couple, b c=L,2 the arm of 
the other. Place the couples so that -f-F 2 and — Fi meet at b. 
It is evident that the resultant of the two forces at b is noth- 
ing, and there remain Fi, — F 2 with an arm a c, which is the 
resultant couple. 

Take any point O in the plane xyz ; draw OMi perpendic- 
ular to the plane of F1L1, proportional in length to its moment 
and so that F1L1 will appear right-handed looking from O 



i5 

toward Mi. Similarly draw OM 2 to represent the couple F 2 L 2 . 
Complete the parallelogram OM,MM 2 ; OM its diagonal will 
represent the resultant couple, since it is perpendicular to the 
plane of the resultant couple and proportional to the resultant 
moment. That is OMi : OM 2 : OM :: F1L1 : F 2 L 2 : FL:: Li : 
L 2 :L. 

The following proof may be preferred : 

39. — Let A, B (Fig. 26) be two couples in the planes whose 
intersection 01 is perpendicular to the plane xyz. Reduce 
the couples to two equivalent couples having a common arm 
OC. Let FiOC and F 2 OC be the equivalent couples. Place 
them so that Fi and F 2 shall intersect at C and Fi and F 2 at O. 
The resultant Ri of Fi and F 2 at C will be parallel to the re- 
sultant R 2 of Fi and F 2 at O, since Fi at C is parallel to Fi at 
O and F 2 at C is parallel to F 2 at O. Hence the resultant cou- 
ple will be R, OC. At the point O draw OMi perpendicular 
to the plane of the couple FiOC, and of length proportional 
to its magnitude. Draw 0M 2 similarly for F 2 OC. Complete 
the parallelogram OM, MM 2 ; OM will represent the resultant 
couple, since OM, MM 2 is OF,R 2 F 2 turned round through 
90 . That is OM, : OM, : OM :: F^; : F 2 L 2 : FL :: F, : F 2 : F. 

40. — Cor. 1. If two couples acting upon a body be represe7ited 
in moments, positions of axes and directions by the two sides of 
a triangle, taken in order, the third side will represent the mo- 
ment, position of the axis and direction of the resultant couple. 

The proof (Fig. 27) is similar to that of §18. 

2. Any number of couples. 
41. — Cor. 2. If all the couples acting upon a body be repre- 
sented in moments, positions of axes and directions by all the 
sides of a polygon taken in order, except one, the excepted side 
will represent the resultant of the given couples in moment, po- 
sition of axis and direction. 

Let M, (Fig. 28) represented by OA be right-handed, look- 
ing from A to O. 

Let M 2 , represented by AB, be right-handed, looking from 
B to A. 



i6 

Let M 3 , represented by BC, be right-handed, looking from 
Cto B. 

Let M 4 . represented by CD, be right-handed, looking from 
D to C. 

Let M 5 , represented by DE, be right-handed, looking from 
E to D. 

Let M 6 be represented by EO. 

Then OE will be the resultant couple, right-handed looking 
from E to O. The polygon may be plane or not plane. 

3. Three couples in different planes. 

42. — Cor. 3. If the three adjacent edges of a parallelopiped 
represent the moments, positions of axes and directions of three cou- 
ples acting upon a body, the diagonal drawn frorn the intersection 
of these three edges, will represent the resultant couple in moment, 
position of axis and direction. This proposition is called the 
" Parallelopiped of Couples." 

Let OA, OB, OC (Fig. 29) represent the three couples. By 
Prop. §38, OD will represent the couple which is the resultant 
of OA and OC. Similarly OM will represent the resultant of 
OD and OB and therefore of OA, OB, OC. 

Let M be the moment of the resultant couple then 
OM*=OA 2 + OB* 4. OC 8 . 



[1]. M^i/M 2 j-f M 2 2 -fM 2 s , which is themagnitude of the mo- 
ment of the resultant couple. 

Let /, /i, v be the angles which OM makes with OA, OB 
and OC. 

2. Then cos. /= — L . 
L J M 

r 1 on, M 2 

[3]. Then cos. /*=--, 

[4]. Then cos. v =^L 3 . 
J M 

The last three equations give the direction of the axis of 
the resultant couple. Hence the above equations determine 
the moment, direction, and position of the axis of the re- 
sultant couple. 



17 

(6). Forces Not in the Same Plane. 

i. Acting at a Point. 

43. — Cor. 6. If three adjacent edges of a parallelopiped repre- 
sent in magnitude and direction three forces acting on a body the 
diagonal of the parallelopiped drawn from the intersection of these 
three edges will represent the resultant in magnitude and direction. 
This proposition is called the " Parallelopiped of Forces." 

For, let OF n OF 2 , OF 3 (Fig. 30), the three edges of a paral- 
lelopiped, represent in magnitude, direction, and point of ap- 
plication the given forces. The resultant of F, and F 2 is R l8 
OR is the resultant of F 3 and R n and therefore of F x , F 2 F 3 . 

Now since OR^OF^+OF^-fOFs 2 , 

[1]. R=VF?+F 2 2 TF7> that is, the magnitude of the re- 
sultant is equal to the square root of the sum of the 
squares of the three components. 

Again, let \ fi, v, be the three angles which the resultant 
OR makes with Ox, Oy, Oz, then 

[2]. cos. 1-=-^' 

r- -1 F 2 

[3]. cos./;-— . 

* F 3 
[4]. cos.v=- R -. 

These equations give the direction of the resultant. 

44. — Prob. Find the resultant of any number of forces at a 
point, not in the same plane, by rectangular co-ordinates. 
See §75. 

2. Not at a Point. 
a. Parallel Forces. 

45. — Prob. Find the resultant of any number of parallel 
forces not in the same plane. See §j6. 

b. Inclined Forces. 
46. — Prob. Find the resultant of any number of inclined 
forces not acting at a point and in different planes. See §78. 



i8 

II. THE PRINCIPLE OF MOMENTS. 

47. — Definition. — The moment of a force or torque is, &c. 
See §27. 

48. — The moment of a force in reference to a point is the 
same as the moment of the force relatively to a line passing- 
through the point and perpendicular to the plane containing 
the line of action of the force and the point. 

49. — Prop. The moment of a force may be represented by 
twice the area of a triangle formed by drawing lines from the 
centre of moments to the extremities of the line representing 
the force. 

Let C, (Fig- 31) be the centre of moments, F the given force, 
CP its arm. Then F ■ CP will be the moment. From C draw- 
lines to A and F. The area of the triangle AFC=^AF • CP 
=y 2 V - CP. Therefore F • CP or the moment is equal to 
2 • area AFC or twice the area of the triangle. 

50. — Prop. The moment of the resultant of any number of 
forces relatively to a point in the plane containing their lines 
of action, equals the algebraic sum of the moments of the com- 
ponents relatively to the same point. This proposition is called 
the " Principle of Moments." 

(a). Forces in a Plane. 

The proof will first be given for two inclined forces. 

51. — Case I. Take the center of moments so that it is not 
inclosed by the lines which represent the two forces. 

Let Fi and F 2 (Fig. 32) be the two given forces and R their 
resultant ; let C be the 'center of moments. From C draw 
perpendiculars to the lines of action of the forces. 

Then R ■ CP=F X ■ CPx+F 2 ■ CP 2 . 

From C draw CA, CB, CD, CE, to the extremities of the 
lines representing the forces. 

Then the area CABD=area CAB+area BCD=area CAD+ 
area ABD=area CAB+area CAE4-area ABD. Since 



J 9 

Area BCD-^BD ■ CP 3 =^BD (CP 2 + P P 3 )=^BD ■ CP 2 + 
>^BD • P 2 P 3 =area CAE+area ^ABDE^area CAE+area ABD 

. • . 2 area CAD=2 area CAB+2 area CAE, that is, the mo- 
ment of the resultant equals the algebraic sum of the moments 
of the components. 

52. — Case II. Take the center of moments so that it is en- 
closed by the lines of action of the forces. 

52a. — Case III. Take the center of moments so that it is 
on the line of action of the resultant. 

53. — Case IV. Prove the proposition for any number of 
inclined forces whether their points of application coincide or 
not. 

In Fig. 33, 

Ra-CPr-F^CPi+F.CP,. 
R.CPr=R 1 CPr 1 + F 3 -CP 3 . 
. • . R CP r =F! • CPi+F 2 • CP 2 +F 3 • CP 3 . 

54. — Case V. Prove the proposition for two parallel forces. 

1. When the center of moments is between the lines of ac- 
tion of the two forces. 

Let Fi and F 2 (Fig. 34) be two parallel forces and R their resul- 
tant. Let C be the point about which the moments are to be 
determined. The moment of F x relatively to C is Fi ■ A X C— F, 
(A, A- AC)=F X • AiA— F, • AC. Similarly the moment of F 2 
relatively to C is — F 2 ■ A 2 C= — F 2 (A 2 A+ AC) = — F a ■ A 2 A— 
F 2 ■ AC. Their sum is Fi • AiC— F 2 • A a C=Fi ■ AiA— Fi ■ AC 
-F 2 • A 2 A — F 2 AC = — Fi ■ AC — F 2 - AC = - (F,+ F a ) AC, 
which equals the moment of the resultant. 

2. Let C (Fig. 35) be beyond A,. 

3. Let the forces be opposite in direction. 

4. Let the forces form a couple. 

55. — Case VI. Prove the proposition for any number of 
parallel forces. (Fig. 35a.) 

56. — 5. Prove the proposition for any number of forces in 
the same line. 



20 



B. RESOLUTION. 



57. — Definition. — Resolution is the process of finding the 
components of a given force. 

/. By the ' i Parallelogram of Forces. ' ' 

58. — Prob. I. Resolve a given force into two components. 

Let AR (Fig. 36) be the given force ; on AR construct any 
triangle AF a R. Complete the parallelogram AFjRF,. Then 
AF X and AF 2 are components of R, since they are the two ad- 
jacent sides of a parallelogram the diagonal of which repre- 
sents their resultant, which is the given force. §17. 

It is evident that any other triangle on AR would fulfil the 
conditions ; hence a force may be resolved into an infinite 
number of pairs of components.. 

59. — Prob. II. Resolve a given force into any number of 
components. 

Resolve each force into its components and each component 
into its components. 

60. — Prob. III. Resolve a given force into two components 
at right angles to each other. 

Let R (Fig. 37) be the given force ; construct on AR a right 
triangle AF X R ; complete the rectangle AF T RF 2 . It is evident 
that AFj and AF 2 at right angles to each other are the re- 
quired components, also that 

AF 2 =AR cos. RAF 2 . 

AF=ARsin.RAF 2 . 

61. — Prob IV. Resolve a given force into two or more par- 
allel forces. 

62. — Prob. V. Resolve a given force into three components 
acting along three rectangular axes. 

Let O (Fig. 38) be the point of application of the given 
force OR. Through this point pass three rectangular planes 
xy, xz, yz. From the extremity of the line representing the 
force draw perpendiculars to the planes ; RPj to zy, RP 2 to 



21 

zx, RP 3 to xy and complete the parallelopiped. It is evi- 
dent that F„ F 2 and F 3 are the required forces, since they are 
the three adjacent edges of the parallelopiped whose diagonal 
is the given force. See §43. 

If A, ^ v are the three angles which R makes with the three 
rectangular axes, 

V = R cos. /C ; F 2 =R cos. fi ; F 3 =R cos. v. 

63. — Prob. VI. Find the components of a given force, along 
and at right angles to any line inclined to the direction of the 
force. 

Let OFi (Fig. 39) making angles a n /9 n y x with Ox, Oy, Oz, be 
the given force and OL making angles « 2 , /3 2 , ^ 2 with Ox, Oy, 
Oz, the given line. Then by the ' 'parallelogram of forces, ' ' OF 2 
and FjF.2 are the required components. Let the angle F ] OF 2 
equal d. Then OF 9 =F 1 cos 6. 

But, cos. #— cos. a x cos. « 2 +cos. & cos. /9 2 -(-cos. y 1 cos. y v 
. ' . OF.^F, cos.ftj cos.a.+FjCOS. ^ cos. /? a +F 1 cos.?' 1 cos.?' 2 
. • . OF 2 =X x cos. « 2 +Y, cos. ^ 9 +Z 1 cos. y^ if Xi, Yi, Z x are 
the components of F l along Ox, Oy, Oz. 

Hence, the component of a given force along ajiy line inclined 
to its direction is the sum of the projections of its rectangular 
components on that line. 

Also F.F^OFj sin. d, which is the component perpendicu- 
lar to the given line. 

2. By Couples. 

64. — Prop. I. A force may be replaced by an equal and 
parallel force in the same direction at a7iy other point in its 
plane and a couple whose moment equals the moment of the 
original force about the point. 

Let the force Fi (Fig. 40) act at Ai. Take a new point O 
in the plane of Fi and in Ox, perpendicular to the same plane. 
F x may be replaced by V\ at O, equal to, parallel to, and in the 
same direction as F n and a couple Fi- L x whose moment is the 
moment of Fj in reference to O. For, at O apply two oppo- 



22 



site forces F'i and — F'„ equal to each other and to F 1# They have 
no resultant. Hence Fi at A 1 is equivalent to F^ at O and a 
couple FjLi around Ox. 

65. — Prop. II. A force may be replaced by a force and two 
couples, whose planes are at right angles to each other. 

Let Fi (Fig. 41) at Ai be the given force. Draw AiB paral- 
lel to OC. At B apply two opposite forces Fi and — F 2 each 
equal and parallel to F, at Ai. At O apply two opposite for- 
ces F 2 and — F 2 , each equal and parallel to Fi at A x . The ap- 
plication of these forces does not change the effect of Fi at A n 
since their resultants equal zero. 

Hence, Fi at A 1 is equivalent to F a (=F 1 ) at O, and a couple 
Fi, — F, with an arm AjB, right-handed looking from x toward 
O around Ox, and a couple F l5 — F 2 (=F,) with an arm OB 
left handed around Oy looking from — y towards O. 

66. — Prop. III. A force may be replaced by a single force and 
three couples in planes at right angles to each other. 

Let Fi (Fig. 42) be the given force. Resolve it into three rec- 
tangular components Xi, Y n Z n parallel to three assumed planes 
zy, xy, zx. Call the arms of couples parallel to each axis,' 
x l5 y x , ?,. 

Now by the preceding propositions, 



Fr 



I Z x ^> 



Single 

Force. 

~ At 


" x t 


COUPLE. 


Around 
Ox 


Around 
Oy 


Around 
Oz 




+x lZl 


-x lYl 


Y, 


-Y,z, 


+,y lXl 


Zx 


+Zj, — Z,x, 





Now, finding the resultant of X n Y ]? Z, which we will call 
F (1), and the resultant of the couples in the same plane, 
around Ox, \Z x y x — Y^Jte); around Oy, [X^ — Z 1 x 1 ](3); around 
Oz, [YjXj — X^J (4), we have the original force equivalent to 
an equal and parallel force in the same direction and the three 
couples represented by (2), (3), (4). 



23 



In naming the direction of the couple always look from the 
extremity of the axis towards the origin. 

67. — An examination of the preceding propositions shows 
that a single force may always be replaced by an equal force 
and a couple; for in the second proposition the two couples in 
planes at right angles may be reduced to a single one by the 
" parallelogram of couples" and in the next proposition the 
three couples may be similarly reduced by the " parallelopiped 
of couples." 

COMPOSITION OF FORCES. (Continued from Sec. 46.) 
68. — Prop. I. The resultant of a force a?id a couple in the same 
plane is a force, equal, parallel to and in the same directio7i 
as the original force, btit removed from it a distance equal to 
the moment of the couple divided by the force. 

Let Fi (Fig. 43) be the given force and F 2 L 2 the given cou- 
ple. Reduce the couple to an equivalent couple having the 

force F x (§33*), V*^= Rx ; x = ?^. 

-Tl 

Now, apply the single force to the couple as in Fig. 44. 
Fi, — Fj have no resultant, and there will remain F, equal 
to the original force but moved parallel to itself a distance 
F 2 

x = -— 

F, 
69. — Prop. II. Find the resultant of any number of forces, 

at a point, in the same plane, by rectangular co-ordinates. 

Let Fi, F 2 , F 3 (Fig. 45) making angles a h a 2 , a s with the 

axis Ox be the given forces. Draw the axis Oy perpendicular 

to Ox. Resolve each force into components acting along these 

two axes ; then by §60, 



U 



F 2 = 
F 3 : 



Call resultants 



Along Ox. 


Along Oy. 


X x = Fj cos. a, 


Y, = F : sin. a\ 


X, = F 2 cos. a. 2 


Y 2 = F 2 sin. a 2 


X 3 = F 3 cos. as 


Y 3 = F 3 sin. a s 


! x 


Y 



24 



Now, find the resultant of the equivalent forces by §22. 
X=Xi-fX 2 +X 3 =F 1 cos. #i+F a cos. # 2 -f-F 3 cos. a s 
Y=Y 1 +Y 2 +Y 3 =F 1 sin. #i+F 2 sin. # 2 -f F 3 sin. a s . 

Construct a new figure with X and Y as adjacent sides 
(Fig. 46). Complete the rectangle ; its diagonal will be the 
resultant required. 



[1]. R = 1/X 2 + Y«=i/(F 1 cos. « X + F 2 cos. « 2 -f F 3 cos. a z f 
+ (F X sin. a x -\- F 2 sin. # 2 + F 3 sin. # 3 ) 2 , which is the magnitude 
of the resultant. 

L,et / be the angle ROX then 

, X . ,* Y ,' Y ■ X 

3. / = — ; sin. A = — ; tan. l— — \ cot. A = — ; any one 



W- 



Y 

R ;tan ' X 



R' R' X' Y 

of which equations will give the direction of the resultant. 

70. — Prop. III. Find the resultant of any number of par- 
allel forces in a plane. 

Let Fi, F 2 , F 3 , F 4 (Fig. 47) be the given parallel forces. At 
any point O erect an axis Oz perpendicular to the plane of the 
forces. Let OAi=xi ; OA 2 =x 2 ; OA 3 =x 3 ; OA 4 =x 4 . 

Then, by §64. 



— F 1 at AiO 

F 2 at A =0 

-F 3 at A s o 

F 4 at AiO 

Call the resultants 



Adding the above columns according to §§22 and 37, 

[1]. R = — Fi+F 2 — F 3 +F 4 which gives the magnitude of 
the resultant. 

[2]. Rx=Fix 1 +F 3 x 3 — F 2 x 2 — F4X4. 



Force at 


Couple Around Qz 


— Y x 

+F 2 
-F 3 
+F 4 


+F lXl 

F,x 2 

+F 3 x 3 

— F4X4 


R 


Rx 



x= 



FiXj+FsXs — F 2 x 2 — F 4 x 



F, + F 2 — F 3 +F 4 



- , which gives the arm 



25 

of the resultant couple or the perpendicular distance of the line 
of action of the resultant from the origin of co-ordinates. 

[3]. The direction is given by the sign of R. 

The direct application of the "principle of moments" 
would have given x. §50. 

71. — Prop. IV. Find the resultant of any number of inclined 
forces, not having the same point of application, in a plane, by the 
graphical method. 

Let Fi at Ai, F 2 at A 2 , F 3 at A 3 , and F 4 (Fig. 48) at O be the 
given forces. At the point O erect an axis Oz perpendicu- 
lar to the plane of the forces. Then, replace each force by an 
equal force at O and a couple around Qz, §64, as follows : 



F, at A,=o 

F 2 at A 2 =o= 
F s at A s =o= 
F 4 at O =0= 

Call resultants 
Find the resultant of the forces at O by the "polygon of 



Force at O. 


Couple around O2. 


Fx 
F 2 
F 3 
F 4 


-F,L, 
-F 2 Ls 
— F3L3 




R 


RL 



forces. 



?23 



Add the moments of the couples to get the resultant mo- 
ment. §37. 

RL = -F 1 L-F 2 L-F 3 F 3 . 



I^-I^L.-F^ 
R 



, which gives the arm of the re- 



sultant couple or the perpendicular distance of the line of ac- 
tion of the resultant from the origin — to determine which the 
" principle of moments" might have been directly applied. 

72. — Prop. V. Find the resultant of any number of forces in 
the same plane by rectangular co-ordinates. 

Let Fj (Fig. 49) at A n whose co-ordinates are x x , y, ; F 2 at A 2 
with co-ordinates x 2 , y 2 ; F 8 at A 3 with co-ordinates x 3 , ys ; &c, 
be the forces, the two rectangular axes being Ox, Oy, with O 



26 

as the origin. Let F n F 2 , F 3 make angles a x , et 2 , a s respec- 
tively with Ox. Resolve each force into components X n Y : ; 
X 2 , Y 2 ; X s , Y 3 , parallel to the assumed axes. Replace each 
component by a single force at O and a couple around the axis 
Oz, which is perpendicular to the plane of the paper at O, and 
tabulate the result as follows : 



m 



F =o 



X 

Y 



n y 3 = 



Call resultants 



Single Force at 0. 


Couple 

Around 

Oz. 

X.7, 

x 2 y 

-Y 2 x 2 

— X 3 y 3 

— Y 3 xs 


Equivalent Couples. 


Along Ox. 
X 2 =Fj cos. a 1 

X =F 2 cos. a % 

X 3 =F 3 cos. a z 


Along Oy. 


Y x — F, sin. a 1 

— Y 2 =F. 2 sin. a 2 

Y 3 =F 3 sin. a z 


}=C=[X iy -Y lXl ] 
}=c=LX,y-Y 2 xJ 

W[-X 8 y 8 -Y 8 x s ] 

1 


X 


Y 


Mz 





Now, find the resultant of the forces acting along Ox. 
X=X 1 +X 2 +X 3 ~Fi cos. tfj+F 2 cos. <? 2 +F 3 cos. a 3 
Also, the resultant of the forces acting along Oy. 
Y=Y,— Y 2 +Ys=F, sin. a— F 2 sin. « 2 -|-F 3 sin. a s . 

Construct a figure (Fig. 50) with X and Yas adjacent sides. 
Complete the rectangle and its diagonal R drawn from O' will 
be the resultant. 



[1]. R— }/X 2 -f-Y 2 , which gives the magnitude of the re- 
sultant. 



Let / represent the angle which R makes with O'x. 
X 
R 



[2]. Then cos. 



sin. /= — ; either of which equations will give 

the direction of the resultant. 

The perpendicular distance of R from is still required. 
In the last column, the couples all act around the axis Oz. 



2 7 

By §37, the resultant moment may be found. 
M z ^RL=Xi yi — Y^-f-X.y,— Y 2 x a — X 3 y 3 — Y 3 x 3 . 

ri T X,y t — Y i x ! + X a y s - Y 2 x 2 — X 3 y 3 — Y 3 x 3 , 

[3]. L= 7" — ■ , which 

L ^ J l/X 2 + Y 2 

gives the required distance. 

73. — The u principle of moments" might have been direct- 
ly applied as follows : 

Let F n F„ F 3 (Fig. 51) with points of application A x , A 2 , A 3 , 
whose co-ordinates are x x , y x ; x 2 y 2 ; x 3 y 3 ; making the angles 
<2 n # 2 , <3!, with Ox, be the given forces. Taking the origin as 
the center of moments, let fall upon the lines of action of the 
forces the perpendiculars Li, L 2 , L 3 . Resolve each force into 
two components parallel to the rectangular axes Ox, Oy. 
These components are represented by X>, Y,; X 2 , Y 2 ; X 3 ,Y 3 . 
Call the resultant R and its unknown arm L. Then by the 
u principle of moments," the moment of the resultant, 

_ RL=P 1 L 1 +F.^ + F.^. = [X 1 y 1 -Y;xJ+ [X i y,-Y,xj+ 

[— X 3 y 3 — Y 3 x 3 ] = (F 1 cos. aj 1 —. F x sin. a 1 x 1 ) -f (F 2 cos. # 2 y 2 — 
F 2 sin.<2 2 x 2 )+( — F 3 cos. a 3 y 3 — F 3 sin. <7 3 x 3 )=F 1 (y 1 cos. ^-J-Xj sin. 
<?i) + F 2 (y 2 cos. tf 2 -f x 2 sin. # 2 )— F 3 (y 3 cos. rf 3 -f-x 3 sin. a z ). 

Hence, L=F 1 (y 1 cos. a 1 — xisin. tfJ+F, (y 2 cos. <z 2 -f-x 2 sin.# 2 ) 

— F, (v, cos. a } -f- x 3 sin. aS) ,. , . ,-, . , , • , 
3 -^- 3 lJ — ? li } which is the required distance 

of the ///z<? of action of the resultant from the origin. 

74. — To find the equation of the line of action of the re- 
sultant, construct a new diagram, Fig. 52, in which 
X = X, + X 2 4- X 3 and Y = Yi — Y 2 + Y,. 

Complete the rectangle, its diagonal will be the resultant. 
Let x and y be the co-ordinates of the point of application of 
the resultant. Then by the " principle of moments" 

RL = +Xy — Yx. 

But X = R. cos. /; Y = R sin. /, 

Hence RL = + R cos. / y — R sin. / x. 

L = -f- y cos. A— x sin. /, / being the angle which the re- 



28 



sultant makes with Ox. Hence the arm of the resultant, &c. 

75. — Prop. VI. Find the resultant of any number of in- 
clined forces whatever, at a point. 

Let F n F 2 , F 3 (Fig. 53) be given in magnitude and direc- 
tion. Assume the three rectangular axes Ox, Oy, Oz. 

Let F x make the angles a iy ft, y 1 with Ox, Oy, Oz respec- 
tively, and let F 2 and F 3 similarly make the angles <? 2 , ft, Tit 

*„ &, r%- 

Then, 





Alojig Ox. 


Along Oy. 


Along Oz. 


ViO 


X, = Fj cos. #i 


Y 1 = Fi cos. ft 


Z, = F, cos. r, 


F 2 =o^ 


X 2 = F 5 cos. a 2 


Y 2 = F 2 cos. ft 


Z 2 = F 2 cos.^ 2 


F 3 =o 


— X 3 =F 3 cos. ^ 3 


Y 3 = F 3 cos. ft 


-Z 3 =F 3 cos. ? ' s , 


Call resultants 


X 


Y 


Z 



The resultant of Fi, F 2 , F 3 will evidently be the same as 
the resultant of the equivalent forces. 

X— X 1 -f-X 2 — X 3 =Fi cos. ^ X +F 2 cos. a 2 — F 3 cos. a s . 

Y=Y!+Y a 4 Y 8 =F 1 cos. ft+F 2 cos. ft+F 3 cos. ft. 

Z=Z 1 +Z 2 -Z 3 =Fi cos. r,-f F 2 cos. r — F 3 cos. r% . 

Now, from O lay off on Ox, Oy, Oz (Fig. 54) the forces X, 
Y, Z. Complete the parallelopiped, its diagonal will be the 
resultant R. 

[1]. R= l /X Ji +Y 2 +Z 2 =i/(F7cosT fli-r-F, cos. a 3 + F 3 cos. aj* 
-HF, cos. ft+F 2 cos. ft+F 3 cos, ft) 2 + (F, cos. 7*1+ F 3 cos. 7-2 + 
F 3 cos. y 8 ) a , which gives the magnitude of the resultant. 

Let R make the angles /, //., v, with Ox, Oy, Oz respect- 
ively, then, 

[2]. 



COS. / = 



X 



[3]. cos.^ = 
[4]. cos.v — 



2 9 



which three equations give the direction of the resultant. 

76. — Prob. VII. Find the resultant of any number of par- 
allel forces whatever. 

Let F n F 2 , F 3 (Fig. 55) acting at A n A 2 , A 3 be the given 
parallel forces. Assume the rectangular axes Ox, Oy, in a 
plane perpendicular to the lines of action of the forces. Then 
according to the principles already shown, §65, 



Call resultants 

Hence, the resultant of F 1? F 2 , F 3 will be the resultant of 
the three single forces at O. and the six couples — three around 
Ox, three around Oy. 

[1]. R=F 1 +F 2 — F 8 , which gives the magnitude of the re- 
sultant. See §22. 

The co-ordinates of its point of application are determined 
by finding the resultant couples around Ox and Oy. Let x 
and y be the co-ordinates of the point of application. 

[2]. Mx "= R • y = Fi ■ yi + F 2 • y^+F 3 ■ y 8 . 
y ==F ry 1 + F 2 -y 2 + F 3 -y 3 
' F.+F.+E, " 

[3]. M y = R • x =-F, . x _F 2 • x 2 _F 3 ■ x 3 . 

■F„ • x — F 3 • x a 



Single Force 
at O. 


COUPLES 


Around Ox. 


Around Oy. 


F 1 
-F. 


+ F.-Y, 
+ F 5 -y s 

+F s -y 3 


-F, ■ x, 

-T 2 ■ X 2 

— F 8 • x a 


R 


,M X 


M y 



- -F, 

x = - 



Xr 



F.+F.+F3 

The resultant of M x aud M y may be determined by con- 
structing a rectangle (Fig. 56) whose adjacent sides represent 
the moments of the resultant couple around Ox and Oy. 

The diagonal M will represent the resultant couple. 

(1) M=i/M^ + M2 . 

Let I be the angle which M makes with Ox, then 



30 



(2) 



cos. X = ™* 
M 

Mv. 



sin. A 



M 



either of which equations will determine the direction of the 
axis of the resultant couple. 

Hence, the resulta7it of any number of parallel forces may 
be a single force, and a couple whose axis is perpendicular to 
the line of action of the given force. 

yj. — Cor. If R equals zero, there will be a resultant couple, 
which may be determined as above. 

78. — Prob. VIII. Find the resultant of any number of forces 
whatever. This proposition shows that any system of forces, 
acting upon any system of points in a rigid body, is eqttiva- 
lent to a single force or a couple — or {excepting parallel forces) 
a single force and a couple whose axis is parallel to the line of 
action of the force. 

Let Fi (Fig. 57) making angles a n /?i, y x with three assumed 
rectangular axes Ox, Oy, Oz, and F 2 making angles « 2 , $*, yi 
with the same axes, be the given forces. 

Resolve each force into three components parallel to the 
three axes — and each of these components into a single force 
at O and two couples around their respective axes. Tabulate 
the values as follows : 




Call resultants 



Single Forces at 0. 


Coiiples. 


Along 
Ox 


Along 
Oy 


Along 
Oz 


Around 
Ox 


Around 
Oy 


Around 
Oz 


X^FjCos.^ 
X 2 =F 2 cos.« 2 


Y^F^os./?, 
Y 2 — F 2 cos./? 2 


Z 1 =F 1 cos.^ 1 
Z 2 =F 2 cos.^ 2 


-Y lZl 
z.y, 

Y,z 2 

-z 2 y 2 


+x lZl 

— Z lXl 

-X 2 z 2 
Z 2 x 2 


-x lYl 

Y, X , 

-X 2 y 2 
Y 2 x 2 


X 


Y 


Z 


M x 


M y 


M z 



3i 

Now, find the resultants of the single forces along Ox, Oy, 
Oz, and finally the resultant of these resultants. 

X = X, + X 2 = F x cos. a x + F 2 cos. « 2 . 

Y = Y, + Y 2 = F x cos. ft + F 2 cos. ft. 

Z - Z x + Z 2 = F, cos. Tl + F 2 cos. r 2 . 

Construct a new parallelopiped (Fig. 58) whose adjacent 
edges are X, Y, Z ; draw the diagonal, it will be the result- 
ant, R. 



[1] R=VX*-j-Y2+Z 2 = j/fF, cos. « +F 2 cos. « 2 )2-f(F lC os. ft 



+ F 2 cos. /5 2 ) 2 +(F! cos. f, + F 9 cos - Tif, which gives the magni- 
tude of the resultant. 

be the angles which R makes with Ox, Oy, 



Let /I, ft, 


*> 


be 


Oz, th 


en 






M 


cos. 


; k = 


X 

"R 


[3] 


cos. 


^= 


Y 
~R 


[4] 


cos. 


v = 


Z 
"R 



These three equations give the direction of the resultant. 

Find, also, the resultant moment around the three axes and 
by the u parallelopiped of couples" the final resultant moment. 

The moment of the resultant couple around Ox is 

M x =[Z iyi -Y lZl ] + [— Z 2 y 2 +Y 2 z 2 ]=[Fi cos. nyi-FiCOS. ftzj 
+ [— F 2 cos. ?- 2 y 2 + F 2 cos. ftz 2 ] = F, (yi cos. y x — z, cos. ft) + 
F 2 (— y 2 cos. r 2 +z 2 cos. ft). 

The moment of the resultant couple around Oy is 

M y = [4-X^ — Z x xi] + [— X 2 z 2 + Z 2 x 2 ] = Fi (z, cos. a x — Xi 
cos. X\) — F 2 (z 2 cos. « 2 — x 2 cos. f 2 ). 

The moment of the resultant couple around Oz is 
M z =[Y lXl - X,yJ + [Y 2 x 2 — XtfJ-Fifo cos. ft— yi cos.aj 
+F 2 [x 2 cos. ft— y 2 cos. aj. 

Now, construct a new parallelopiped whose three adjacent 



32 

sides represent M x , M y , M z ; its diagonal will represent the 
moment^ direction and position of the axis of the resultant 
couple, which call M. Then 

[i]. M=i/M| -f M* + Mf , which gives the moment of the 
resultant couple . 

Let £, C, ">?> be the angles which M makes with Ox, Oy, Oz, 
then 

[2]. cos.6=-- 

r 1 r M ^ 

[3]. cos.C= M - 

These equations give the direction of the axis of the resultant 
couple. 

79. — As indicated in (Fig. 58), R has been found in magni- 
tude and direction ; also M in moment, position of the axis, 
and direction. R and M may make any angle d with each 
other. 

[1]. It is obvious, that if the axis of M makes an angle 
with the direction of R. 

COS. 6 = COS. X COS. f -[- COS. /J. COS. C + COS. V COS. 7j. 

To reduce R and M to a simpler form, resolve M into two 
components, one M P = M sin. #, around an axis perpendicular 
to R and in the plane of R and the axis of M. This couple 
(Fig. 59) and R being in the same plane are by §68 equivalent 

^ , ,.'..■,. r Msin 

to R whose perpendicular distance from O is h = — - — > 

R 

Another M r = M cos. 0, whose axis is parallel to R (Fig. 60). 
This case cannot be further reduced. 

[2]. If R is perpendicular to M ; 0=go°. cos. 9o°=o, 
cos. X cos. s-f-cos. /1 cos. £+cos. v cos. q=o. 

XM X Y M y Z M*_ 

r" M r' M + r' M ~ 

X ■ M x + Y • M y 4- Z • M z = o. 



3 3 
These, as above by §68, are equivalent to a single force whose 

T M 

arm L = -- • 
R 

[3]. If R is parallel to M, d=o. cos. 0=1. 

cos. / cos. £ -f cos. a cos. J" -{-cos. v cos. -^ =1. 

/ = S /£ = f V = if 

*-= — % /* = — : v=—vj 

This case admits of no further reduction. 

[4]. If R = o, then the resultant is the couple M. 

[5]. If the moment of the resultant couple M = o, the re- 
sultant is the single force R. 

From this general proposition all the preceding proposi- 
tions mav be deduced. 



C. BALANCE OR EQUILIBRIUM. 

80. — Two or more forces are bala need when they produce no 
change in the rest or motion of the body upon which they 
act. 

It is obvious, then, that the resultant of a balanced system of 
forces must be zero. See §10. 

If the forces act at a point the only condition of equilibrium 
is, therefore, that the resultant force shall equal nothing. 
See §89. This can always be brought about by applying to 
each force in the system an equal and opposite force or by-ap- 
plying a single force, equal and opposite to the resultant of the 
system. 

BALANCE BY THE "PARALLELOGRAM OF FORCES" 
(a). Forces in a Plane. 
1. At a Point. 
81. — Prop. I. A single force cannot produce equilibrium. §10. 
82. — Prop. II. Two forces in the same straight line balance 
each other when they are equal and opposite ; for then their re- 
sultant is nothing. See §21. 



34 

83. — Prop. III. Any number of forces in the same straight 
line will balance when their algebraic sum equals zero. See §22. 

84. — Prop. IV. Two inclined forces at a point cannot bal- 
ance each other si?ice their resultant has value. They can only 
be balanced by a third force equal and opposite to their re- 
sultant. See §17. 

85. — Prop. V. If three forces acting upon a particle be rep- 
resented in magnitude aitd direction by the three sides of any 
triangle taken in order, the forces will balance. This propo- 
sition is called the " Triangle of Forces." 

Let F n F 2 (Fig. 61) be two given forces ; complete the par- 
allelogram OF^Fg and draw its diagonal R. The diagonal 
will represent the resultant in magnitude and direction. Since 
F X R— AF 2 ; in the triangle AF 2 R, AF x represents F n F X R rep- 
resents F 2 and AR the resultant, to balance which a force RA 
must be applied opposite in direction. Hence when balance 
is established, F„ F 2 , RA, taken in order, represent the three 
forces. 

86. — Cor. If three forces balance they must act at a point 
and each must be proportional to the sine of the angle included 
between the direction of the other two. 

Let F n F 2 , F 3 (Fig. 62) be the given forces. It is obvious 
that F 2 must be equal and opposite to the resultant of F 1 and 
F 3 , otherwise their resultant would not equal zero. In the 
triangle ARF„ AF„ F X R and RA represent the forces, hence, 

P, : F 2 : F 3 :: sin. ARF, : sin. RF,A : sin. RAF,. 

F, : F 2 : F 3 :: sin. F 3 AF 2 : sin. F 8 AF l : sin. F.AF^ 

87. — Prop. VI. Any number of forces acting at a point balance 
when represented in magnitude and direction by all the sides of a poly- 
gon taken in order. See § 23. This proposition is called the 
" Polygon of Forces." 

The forces F n F 2 , F 3 (Fig. 63) may be represented by AF X> 
F^,, R X R and the resultant by AR. Hence, to balance F xJ 
F 9 , F 3 , balance R by applying a force RA opposite in direction 
and equal to R, but in so doing the forces are represented by 
all the sides of the polygon taken in order. 



35 

88. — Cor. If the method of rectangular axes is employed, 
then, since R = o ; R = j/X 2 + Y 2 = o, but then X = o ; 
Y = o, that is, the algebraic sum of the resolved parts of the for- 
ces along any two axes at right angles to each other in the plane of 
the forces must equal zero. §69. 

2. Forces not at a point. 

89. — If the forces act upon a rigid body at different points, 
since each force has been proved equivalent to a single force 
and a couple, the conditions of equilibrium are that (1) the re- 
sultant of the single forces shall equal zero and (2) the resul- 
tant couple shall equal zero. 

90. — Prop. I. Any number of inclined forces in a plane 
balance when the resultant, R == zero, a7td the mome7it of R, 
(RL) relatively to every point in the plane equals zero. §72. 

91. — Prop. II. Any number of inclined forces in a plane 
when the method of rectangular axes is used, balance when 
the sum of the resolved parts of the forces along two rectangu- 
lar axes equals zero and the moment of the resultant couple 
about an axis perpendicular to the plane of the forces equal 
zero or 



[1]. R = 1/X 2 + Y 2 or X = o; Y = o. 
[2]. M= sum of (Yx— Xy) =0. 

3. Parallel forces. 

92. — Prop. I. Two parallel forces cannot balance. 

93. — Prop. II. Three parallel forces balance, when each is 
equal and opposite to the resultant of the other two; that is (1) 
each must equal the algebraic sum of the other two. (2) Each 
must divide the distances between the other two inversely as 
their magnitudes. 

Let Fi, F 2 (Fig. 64) be two parallel forces to be balanced by 
a third force F 3 . Find the resultant of F n and F 2 . Balance 
the resultant by applying an equal and opposite force F 3 . 

Since F 3 is equal to R,it must be equal to the sum of F 1 and 
F 2 and since it must be opposite to R, its point of application 
must divide the distance between A, and A 2 inversely as the 
magnitudes of F, and F 9 . §24. 



36 

94- — Prop. III. Any number of parallel forces balance when 
their resultant R,=o and the sum of the moments about every point 
in their plane equals zero. §76. 

(b.) Couples. 

1. With the same axis. 

95. — Prop. I. Two couples with the same axis balance each other 
when of equal moment and opposite in direction, for then, their re- 
sultant moment is nothing. §35. 

96. — Prop. II. Any number of couples with the same axis will 
balance when the algebraic sum of their moments equals zero; for then, 
the sum of the moments of the right handed couples equals 
the sum of the moments of the left handed couples. §37. 

97. — Prop. III. A force and a couple in the same plane cannot 
balance, since their resultant always has value. §68. 

2. With different axes. 

98. — Prop. I. Two couples in planes inclined to each other can- 
not balance each other, since their resultant has value. §38. 

They can only be balanced by a third couple, equal in mo- 
ment and opposite in direction to their resultant ; hence 

99. — Prop. II. If three couples be represented in moments, posi- 
tions of axes and directions by the three sides of any triangle taken in 
order, the couples will balance. §40. This proposition is called 
the u Triangle of Couples." 

100. — Prop. III. Any number of couples with different axes bal- 
ance when represented by all the sides of a polygon taken in order. 
§41. This proposition is called the " Polygon of Couples. " 

101. — Cor. If the method of rectangular axes is employed, the 
sum of the resolved moments along three rectangular axes must equal 
zero. 

[1]. M =0; that is, M x == o; M y = o ; M z = o. 

(c. ) Forces not in the same plane. 

1. At a point. 

102. — Prop. I. Three forces at a point not in the same plane 
cannot balance, since their resultant has value. §43. They 



37 

can be balanced by applying a force equal and opposite to 
their resultant. 

103. — Prop. II. Any number of inclined forces at a point not 
in the same plane balance when their resultant equals zero, that is, §75, 



R = /XHYHZ 2 -o,orX = o; Y=o; Z=o, that is, the 
sum of the resolved parts along three rectangular axes must equal 
zero. 

2. Not at a point. 

(a). Parallel Forces. 

104. — Prop. 1. Any number of parallel forces whatever balance 
when their resultant equals zero, and the sum of the moments in refer- 
ence to two rectangular axes equals zero. §76 

[1]. R = o. ~ X — o. — Y — o. 

[2]. Sum of (yF) = o. Sum of (xF) = o. 

(b). Any number of Forces. 

105. — Prop. I. Any number of forces whatever balance when 
their resultant equals zero and the moment of the resultant couple 
equals zero, that is, 

[1]. R = o,orX = o; Y = o ; Z = o. 

[2]. M = o, or M x = o ; M y = o ; M z = o. §78 



ILLUSTRATIONS OF THE PRINCIPLE OF THE "PARALLELOGRAM OF FORCES" AND 
OF THE "PRINCIPLE OF MOMENTS." 

A. Centre of Mass — Centre of Gravity. 

106. — Definition. Among any number of particles there is 
a point, the distance of which from any plane is the mean of 
all the distances of all the points from that plane. This point 
is called the point of mean position or the me 'an point ox- centre 
of mass of the particles. 



38 

Let A, and A 2 (Fig. 65) be two points, and zy a plane. 
Join A x A 2 ; the point P' midway between A x and A 2 is the 
mean point for these two points. From Ai, A 2 and P' draw 
parallel lines to the plane, and let Ai« 1 =x 1 ; A 2 <7 2 =x 2 ; 
~P'a f = ij. Draw A 2 q 2 perpendicular to Xl . Since the triangles 
A 2 P r Ci and A 2 Aiq 2 are similar. 

A 2 P / :A 2 A 1 ::P / C 1 :A 1 q 2 . 

But A 2 P': AAr.-l : 2. 
.-. P'Q: A x q 2 ::1 : 2. 

P'C == ^ 2 X l X 2 

2 2 

P! „!— — "D/P _L -^ X i X 2 1 X l+ X a 
a — xi—r LxH- x 2 = -i ? -f x 9 = 

2 2 

Again, take a third point A 3 and divide the distance AsP' 
so that 

P'P": P /r A 3 ::l : 2. 
p'P"+p"A 3 =P'A 3 . p/p// ::3 . 1# 

Draw the perpendicular P'q 3 to x 2 , then since P'P r/ C 2 and 
P r A 3 q 3 are similar. 

P'A a :P'P"::A 8 q 3 :,P"C 2 . 
.-. A s q 3 :P"C 2 :: 3 :l. 
p// c _ rr A 3 q 3 _ x 3 — x, 



3 3 

x 3— x, . _ x 1 +x 9 -f-x f 



P"a"= Xl = P"C 9 - r - Xl 

3 
To make the case more general, let 

A.P'iP'Ax::^ : a 2 

A 2 P / +P / A 1 =A 2 A 1 : A 2 P r :: a x +a % \a x 
A 2 A 1 :A 2 P'::A 1 q 2 :P'C 1 
A 1 q f :P , Ci::« 1 +-o,:« 1 

A,q 2 x, — x 2 ^jXj — ^x 2 



P'C^tf 



tf,+tf 2 * «! + ^ 2 ^ + ^3 



Xi r L,+X 2 = : ^X 2 — ■ 

Taking another point P", it may be shown that 

__tfiX 1 -r-tf 2 x 2 -r-<? 3 x 3 
x 2 — : — — 



39 

It is obvious, that it is not necessary that the lines A^, A 2 « 2 , 
A 3 « s , etc., should be drawn perpendicular to the plane, as 
long as they are parallel to each other; also that all the points 
are not necessarily on one side of the plane. 

Hence, the distance of the mean point may be obtained by 
multiplying the distances x n x 2 , x 3 , by the multipliers %, « 2> 
« 3 and dividing the sum of the product by the sum of the 
multipliers. 

107. — By referring to § 76, it is seen that the point through 
which the resultant of parallel forces passes was obtained in 
exactly the same manner. The mean point, then, is a gen- 
eral term, under which the center of parallel forces may be 
regarded as a special case. 

108. — Centre of Gravity. The force of gravity acts upon 
every particle of matter. 

Its direction is towards the centre of the earth. The dis- 
tance from the surface of the earth to its centre is so great, 
compared with the distance apart of the particles of any body 
considered in mechanics (upon its surface) that the forces ex- 
erted upon the body may be regarded as parallel. 

The point of application of the resultant of these parallel 
forces of gravity, is called the center of gravity of the body. 
Hence, finding the center of gravity is simply finding the 
point of application of the resultant of parallel forces. 

It follows from the principles of equilibrium already estab- 
lished that the centre of gravity is that point in a body which, 
if supported, the body will remain at rest in any position. §80 

It is obvious, that, the mean point of a body only coincides 
with its centre of gravity, upon the supposition of parallelism 
of the forces of gravity. 

109. — Prop. I. The centre of gravity of a uniform straight 
line, §4, is at its middle point. 

Suppose the straight line A B (Fig. 66) to be made up of 
an infinite number of uniform particles placed side by side, at 
equal distances from each other. Represent the weight of 
each particle, by a line drawn from the particle, in the direc- 
tion of gravity and proportional to its weight. 



4o 

Find the point of application of the resultant of these forces, 
§25, by taking them two and two at equal distances, on oppo- 
site sides of the centre. The centre of gravity will be in the 
middle of the line. 

no. — Prop. II. Find the centre of gravity of the perimeter 
of a polygon. 

The centre of gravity of each side of the polygon (Fig. 67) 
is at the middle point, §109, that is, the resultant weight of 
each side acts at the middle point. 

Join the centres of gravity, §25, gj and g 2 and let g 6 be their 
centre of gravity. 

The weight of AB : the weight of BC :: g 2 g 6 : g$g v 

AB + BC : BC :: g 2 g* + g 6 gi = g,g, : g 6 g x , which gives g 6 gi. 

Join g 6 and g 3 and let g 7 be the new centre of gravity 

Then, AB+BC : CD :: g 7 g s : g 7 g & . 

AB+BC + CD : CD : g 7 g B -+- g 7 g 6 = g 6 g^ : g 7 g„ which gives 

Continue the operation, until g the centre of gravity of the 
whole perimeter is found. 

in. — Prop. III. Find the centre of gravity of a parallelo- 
gram. §25. 

Let ABCD (Fig. 68) be a parallelogram regarded as made 
up of an infinite number of straight lines parallel to AB. §5. 
The centre of gravity of each of these lines is at the middle 
point, §109 ; hence, the centre of gravity of the whole figure 
must be in a line passing through these centres of gravity. 
Similarly ABCD may be regarded as built up of an infinite 
number of straight lines parallel to BC ; hence, its centre of 
gravity must be in another line passing through these centres 
of gravity, that is, at the intersection of the two lines or at g, 
the centre of figure. 

112. — Prop. IV. Find the centre of gravity of a triangle. 

Let ABC (Fig. 69) be a given triangle ; imagine it made up 
of an infinite number of lines parallel to AB. The centre of 
gravity of each line will be in its middle point by §109. The 



4i 

centre of gravity of the triangle will then be somewhere in 
the line CD, §25, passing through all the middle points of 
these lines. For similar reasons, it will be in the line BE, bi- 
secting the line AC. Hence the centre of gravity, being on 
both the lines BE and DC, must be at their intersection g. 
But where is g in the triangle ? Join the points D and E. The 
line DE will be parallel to BC, for 

AD : AB::l : 2, 

AE : AC:: 1:2. 

AD : AB::AE: AC. 

Since ADE and ABC are similar, 

AD : AB::DE: BC. 

But AD=^AB 

.*. DE=^BC. 

Also, DgE and BgC are similar, 

.'. Cg:gD::BC : DE 

. ' . gD=^Cg=^CD, that is, the centre of gravity of a triangle 
is in a line drawn from the middle point of one of the sides to the oppo- 
site angle, at a distance from the side equal to one third the distance 
of the middle point from the angle. 

113. — Prop. V. Find the centre of gravity of a regular polygon. 

Divide the polygon into triangles. Find the centre of 
gravity of each triangle by §112 and then of all by §25. 

114. — Prop. VI. Find the centre of gravity of two heavy 
bodies. §25. 

Let g 1 and g 2 (Fig. 70) be the centres of gravity of the bodies 
B x and B,. Join gi and g 2 . Then, since the centre of gravity 
is the centre of parallel forces, find the centre of Wi and W 2 
by §25 ; it will be g, so situated that W, : W 2 :: g 2 g : g& or if 
the weights and distance between g, and g 2 are given, by com- 
position, 

W, : Wi -f W 2 :: g„g : g 2 g+g,g = g,g2, all of which terms are 
known except g 2 g which can be found by solving the pro- 
portion. 

115. — Prop. VII. Find the centre of gravity of any number of 
heavy bodies. 



42 

Let B n B 2 , B 3 , B 4 (Fig. 71) be four heavy bodies, whose 
weights are Wi, W 2 , W 8 , W 4 , acting at their respective centres 
of gravity g h g 2 , g 3 , g 4 . Join the centres of gravity g n g 2 of 
B x and B 2 and let their common centre of gravity be g 5 . By 
§25, 

W, : W 2 ::g 2 g c : g x g 5 . 

W 1 +W 9 : W a ::g a g 6 +g 1 g5=g 1 g 9 : g y g 5 . 

Now, suppose B 1 and B 2 concentrated at g 5 . Join g 5 andg 3 
the centre of gravity of B 3 . Let g 6 be the new centre of grav- 
ity, then 

W 1 + W 2 :W 3 ::g 3 g 6 :g 5 g, 

W 1 -fW 2 +W 3 : W 3 :: g 3 g6-r-g5g6=g 5 g 3 : gsg*. 

Similarly W^W.+W.-hW* : W 4 :: g 6 g, : g&i, &c. 

116. — The principle of moments may be applied as follows : 
Assume three planes (Fig. yia) at right angles to each other. 
Let x 3 , x 2 , x 3 , x 4 , be the distances of the centres of gravity of 
the bodies from the plane yz and yi, y 2 , y 3 , y 4 the correspond- 
ing distances from xz and z n z 2 , z a , z 4 from the plane xy. Then 
R . x = W 1 x 1 +W 2 x 2 +W 3 x 3 +W 4 x 4 . 
W jXl H-W 2 x 2 +W 3 x 3 n-W 4 x 4 






w.+w.+w.+w* 

W 1 y I 4-W 1 y 1 +W,y,+ W 4 y 4 



y- 



W 1 H-W 2 +W 3 +W 4 
Wz 2 -fWz 2 + Wz +W 4 z 4 



22 
z — 



w,-t-w 2 +w 3 +w 4 

x, y and z, being the co-ordinates of the point of applica- 
tion of the resultant force, give the position of the centre of 
gravity of the bodies. §76 

116. — Prop. VIII. Find the centre of gravity of a triangu- 
lar pyramid. 

Let ABCD (Fig. 72) be a triangular pyramid. It may be 
regarded as composed of an infinite number of triangular flat 
plates, §6, placed parallel to BCD. Bisect DC atE; join BE; then 
will g x be the centre of gravity of BCD; also the centre of 
gravity of b c d will be at g 2 ; therefore the centre of gravity of 
the pyramid will be in Ag 2 g, ; by similar reasoning it will 






also be in BM. Therefore, it will be at their interseption g. 

Join giM; it is parallel to AB. 

.-. Mg : Mg,::gB : AB. 

Mg: gB :: Mg, : AB :: Egi : EB 1:1:3. 

gB = 3 Mg. 

BM = 4 Mg. 

Mg = y± BM, that is, the centre of gravity of a triangular pyra- 
mid is on the line joining the centime of gravity of any face ivith the 
vertex of the opposite angle, at one quarter the distance from that face 
to the angle. 

117. — Prop. IX. Find the centre of gravity of any pyramid, 
having a plane rectilinear polygon as its base. 

Divide the pyramid into triangular pyramids by taking any 
point in the base and joining it to the angular points of the 
base. The centre of gravity of each is in a plane parallel to 
the base and one-fourth the distance from the vertex. Hence, 
the centre of gravity of the pyramid is in this plane. 

Again, the centre of gravity of the plates parallel to the 
base is in the line drawn from the vertex through the centres 
of gravity of the plates. Hence, the centre of gravity of the 
pyramid is on this line. 

Hence, the centre of gravity of the whole pyramid is at 
one- fourth, &c. 

118, — Prop. X. Find the centre of gravity of a cone. 

119. — Principle of Symmetry. The centre of gravity 
may, sometimes, be found by the principle of symmetry. 

There are certain figures which may be equally divided, so 
that for every point on one side of a line or plane, there is a 
corresponding point on the opposite side of the line or plane; 
the corresponding points being on a line perpendicular to the 
given line or plane. Such bodies are said to be symmetrical 
in reference to the line or plane. 

Take a uniform, straight line (Fig. 73) for example, 

Draw a vertical through the middle point. AB obviously 
fulfills the requirements of the definition. Hence, the weights 
of the particles on the right of the vertical equal the weights 
on the left, and the middle point is the centre of gravity. 



44 

Again, let ABCD, (Fig. 74) be a circular plate. Draw a 
vertical diameter AB. On the right and left sides of AB there 
is a particle, &c, therefore, the centre of gravity must be in 
AB. For similar reasons, it must be in the horizontal diam- 
eter DE, therefore, it must be at the centre. Similar reason- 
ing will apply to rectangles, ellipses, &c. 

Take a uniform sphere. Three planes at right angles to 
each other through the centre will divide the figure symmet- 
rically. Therefore, the centre of gravity will be at their in- 
tersection, that is, at the centre of the sphere. Similar rea- 
soning will apply to circular and elliptic cylinders, ellipsoids, 
parallelopipeds, &c. 

120. — Properties of the Centre of Gravity. — Since the re- 
sultant of all the forces acting on the body must be balanced 
in order to balance the body, the centre of gravity may be de- 
fined as that point in a body or system of bodies which, if 
supported, the body or system will remain at rest in any po- 
sition. §108. 

121. — Prop. I. If a body is suspended from a point, it will 
remain at rest only when its centre of gravity is directly above 
or below the point of suspension. 

Let B (Figs. 75-76) be a body, whose weight is W and cen- 
tre of gravity g, supported by a fixed point O. It is evident 
that if B does move, its centre of gravity will describe the arc 
of a circle whose radius is Og. At any instant the direction 
of motion will be a tangent (gT) to this arc. Is there any 
force that will produce motion along the tangent ? Resolve, 
§58, gW into two components, gT in the direction of the tan- 
gent, and gN perpendicular to the tangent. gN will simply 
pull or push the fixed point O, while gT will produce motion 
along the tangent. Hence, as long as gT has value the body 
will move. 

Now, gT = gW cos. TgW == gW sin. BOz. 

When g is directly above or below O, BOz = ; .'. sin. 
BOz=0 ; and gT = 0. That is, the body will remain at 
rest. 






45 

122. — Hence, the experimental method of finding the cen- 
tre of gravity by suspension. 

Suspend the body and a plumb-bob from a fixed point. 
The centre of gravity will be along the line of the bob. Sus- 
pend them from another point, it will obviously be in this di- 
rection. If the body is not of uniform thickness, suspend 
from a third point. Then, the intersection of these direc- 
tions will give. the position required. 

123. — Definitions. The " line of direction" is a vertical pass- 
ing through the centre of gravity of a body, 

The base of support is the surface enclosed by the extreme 
points of contact of the body with the area upon which it 
stands. 

123^. — Prop. II. If a body is supported upon a horizontal plane \ 
it will remain at rest, when the u line of direction" falls within 
the base of support ; when without, it will fall. 

Let ABCD (Figs. 77-78) be a body resting upon a horizontal 
plane, g its centre of gravity and W its weight. If it move 
at all, g will describe the arc of a circle whose radius is Cg, or 
gD. Resolve gW into two components, one in the direction 
of the tangent to the arc, the other perpendicular to it. It is 
obvious that in the first figure the component is not in the 
proper direction for motion, while in the second it is. Hence, 
in the first case the body will stand and in the second fall. 

124. — Hence, the experimental method of finding the cen- 
tre of gravity by balancing the body across a thin edge. 

125. — Equilibrium. Definition. A body is said to be in a 
condition of stable eqitilibriiim, if, when disturbed, it tends 
to return to its original position. It follows from §§121, 123^, 
that if a body is in a condition of stable equilibrium its centre 
of gravity must rise when the body is disturbed. 

A body is said to be in a condition of Unstable equilibrium, if 
when disturbed it does not tend to return to its original po- 
sition. It follows, from the same propositions, that when a body 
is in a condition of unstable equilibrium its centre of gravity 
must fall, when the body is disturbed. 



4 6 

A body is said to be in a condition of indifferent or neutral 
equilibrium when it remains at rest in any position. It fol- 
lows that the centre of gravity must neither rise nor fall when 
the body is disturbed, if in a condition of neutral equilib- 
rium. 

126. — The same body may have these conditions when in 
different positions, as may be shown by placing the ellipsoid, 
prolate spheroid, oblate spheroid, double cone, &c, on end, 
on the side, on the base. 

127. — The various motions of animals, walking, running, 
trotting, galloping, &c, the art of rope dancing, rising from 
a chair, carrying burdens, illustrate the foregoing principles. 
The ballasting of ships furnishes another illustration. 

128. — Stability. — Definition. A body is said to have sta- 
bility when a great force is required to overthrow it. The 
conditions of stability are : 1, lowness of centre of gravity, 
and 2, having the "line of direction" far within the base of 
support. 

Pyramids, tables, quadrupeds, &c, illustrate these condi- 
tions. 

B. Simple Machines. 

129. — Definitions. Simple Machines are contrivances 
by which a small force may be made to balance a larger one, 
or the point of application of one force may be made to move 
more rapidly than the point of application of the other when 
the simple machine is put in motion. 

The power of a simple machine is the force which is in- 
tended to balance the resistance or weight to be overcome. 

There are six simple machines — the lever, the wheel and 
axle, the pulley, the inclined plane, the screw, the wedge. These 
six may be reduced to three — the lever, the pulley and the in- 
clined plane. 

130. — The lever is a bar, free to turn around a fixed axis. 
The fixed axis is called the fulcrum. The lever may be 
straight or bent. 



47 

There are three classes of levers, determined by the relative 
positions of the points of application of the power, the weight 
and the fulcrum. 

In the first class (Figs. 80, 81, 82, 83, 87), the fulcrum is be- 
tween the power and the weight. The crow-bar as ordinarily 
used, the hand spike, the pump-handle, the see-saw, &c, are 
examples of this class. 

In the second class (Fig. 84), the weight is between the ful- 
crum and the power. A wheel-barrow, an oar, a door shut by 
the hand placed near its outer edge, a crane, are illustrations 
of this class. 

In the third class (Fig. 85), the power is between the fulcrum 
and the weight, as, when a ladder is lifted by putting the foot 
against the lower end and pulling at a point below its centre 
of gravity. Certain bones of animals also illustrate this class. 

131. — Prop. I. The law of balance of the lever is the power 
multiplied by its arm equals the weight multiplied by its arm. 

Let A,AA 2 , (Fig. 79) be a bent lever, with forces F n F 2 , in- 
clined to each other. It is obvious, §85, that the resultant of 
these forces must pass through A, the fulcrum. Taking A as 
the centre of moments — by the "principle of moments," §50, 

RXo = Fi- A^+F,- AP r . 

F AP, 
* F • AP = F • K P or — — . 

Cor. 1. The lever is straight and the forces inclined, as in 
Fig. 80. 

Cor. 2. The lever is bent and the forces parallel, as in Fig. 
81. The claw and hammer illustrate this case. 

Cor. 3. The lever is tilted from its horizontal position and 
the centre of gravity of the bar ?s at the fulcrum. Fig. 82. 

Cor. 4. The lever is straight and the forces parallel. Figs. 
83, 84, 85. The forces being parallel the resultant equals F x +F 2 . 

Cor. 5. The lever is supported at both ends and the weight 
is placed in the middle. (Fig. 86.) The resistance at either 
end may be taken as the power. The bar is straight and the 
forces parallel. 



4 8 

Cor. 6. The lever is straight but its centre of gravity is not 
at the fulcrum. Fig. 87. 

Let Wb be the weight of the lever regarded as concentrated 
at g, its centre of gravity, §108. Then, applying the "princi- 
ple of moments," the centre of moments being taken at the 
fulcrum, F, ■ A, A+W b ■ gA = F 2 • A 2 A. 

Cor. 7. Several forces are in equilibrium about a fulcrum. 
Fig. 88. 

R • = F l • A a A + F a • A.A+F, • A 3 A+F 4 ■ A 4 A + F 5 ■ A 5 A. 

. • . — F, • A X A— F 2 • A 2 A— F 8 • A 3 A=F 4 ■ A 4 A + F 5 - A 5 A, that 
is, the sum of the right handed moments equals the sum of the 
left handed moments. 

132. — The principle of the " parallelogram of forces" might 
have been used to obtain the law of balance. 

Let A, AA 2 (Fig. 89) be a bent lever kept in balance by forces 
F, and F 2 ; let A be the fulcrum. It is obvious that, for equi- 
librium, the resultant of F 1 and F 2 must pass through the ful- 
crum. Produce the lines of action of the forces back until 
they met at A 7 ; from A'A 2 cut off a part proportional to F 2 
and similarly from A'A, a part proportional to F,. Complete 
the parallelogram. 

Then, F, : F 2 :: F 2 R : A'F a :: sin. RA'F 2 : sin. RA'F,. 

Draw the perpendiculars AP, and AP 2 from the fulcrum. 

Then, AP, : AP 2 :: A'A sin. RA'F,: A'A. sin. RA'F 2 ::sin. 
RA'F, : sin. RA'F 2 . 

.-. F i: F 2 ::AP 2 :AP, 

. " . F x • APj == F 2 • AP 2 , which is the same as above. 

133. — The lever of the first class is used as a weighing ma- 
chine ; sometimes the arms are made equal and it is called the 
balance; sometimes the arms are unequal, as in the steelyard. 

The balance is used for weighing, by placing the body in 
one pan and adding standard weights to the other until the 
beam is brought back to a horizontal position. 

To make the case general, let A X AA 2 (Fig. 90) be a balance 
with the fulcrum at A, the centre of gravity at g, and having 



49 

arms A^^I^ and A 2 A=<L 2 , making the angles A.Ax—ai and 
A 2 Ax' — a 2 with xAx'. Let W p , and W P2 be the masses of 
the two pans and W the two masses placed, one in each pan. 
Suppose an additional small mass w is added to the right hand 
pan ; the balance will tilt ; allow it to take the position 
A/AA/, making the angle, A.AA/ == A 2 AA/ = 0, with the 
original position. The centre of gravity will move from g x to 
g 2 and gjAg 2 =#. Call Ag n the distance of the centre of 
gravity below the fulcrum, d. Also, let Wb = weight of the 
beam. Then, by the principle of moments, 

(W Pl +W) AP,+W b ■ g 2 m = (W P2 +W+w) AP 2 . 

AP 1 =AA/ cos. A^AP^AA/cos. (ai+d)=l /l (cos. a x cos. 6— 
sin. a x sin. 6). 

g 2 m = g 2 A sin. g x Ag 2 = d sin. 6. 

AP 2 = AA' 2 cos. A' 2 AP 2 = L 2 cos. (a 2 —d)=L, 2 (cos. a 2 cos. 0-f 
sin. a 2 sin. 0). 

(W Pl +W) L, (cos. a, cos. 6— sin. a x sin. #)+W b d sin. = (W P2 
-f-W-fw) L 2 (cos. # 2 cos 0-f-sin. a 2 sin. #). 

/ cos. . sin. 6>\ sin. 

(W Pl + W)L 2 cos. ^ -—sin. a, -fW b d 



cos. # cos. 01 COS. # 

/TX7 ITITI \ T / C0S ' ^ I • Sm - 6 \ 

(Wp 2 + W + w) L 2 cos. tf 2 s +sin. a 2 

\ cos. cos. 0/ 

L t (W Pl +W)cos.^- L a (W Pa ,+ W+w)cos. a 2 

• - tan ' & •L 1 (W Pl -r-W)sin.^+Iv 2 (W P2 +W+w)sin.^ 2 -W b d 
Now, in the actual balance the masses of the pans are made 
equal, that is, W Pl = W P2 = W p ; the arms are made of equal 
lengths, that is, 1^ = 1^= L, .and a x = # 2 = a ; also W = W. 
Substituting these values in the above equation, we have 

ft = L ( Wp+ W) c os - a ~ MW p +W +w) cos. a 

an ' : ~ L (W p + W) sin. a— L (W p +W-f-w) sin. a—W h d 

_ w L cos. a , . 

. * . tan. = ^-^ — r -7- nT — ; , that is, 

W b d— [2(Wp+W)-f-wJIysm.tf 

the tangent of the angle of displacement from a horizontal 
position, due to a small weight added, when the beam is 
loaded, equals the small weight (w) multiplied by the length 
of the arm (L) and the cosine of the angle which the arm 



5o 

makes with the horizontal, divided by the difference between 
the product of the weight of the beam (W b ) and the distance 
(d) of its centre of gravity below the fulcrum and the contin- 
ued product of the length of the arm (L,), the sine of the an- 
gle (a) the beam makes with a horizontal line, and the sum of 
twice the weight of the pan (W p ) and the load plus the extra 
weight added (w). 

134. — Discussion. — Definition. The sensibility of a balance 
is the angle, 6>, through which the loaded balance turns from 
its horizontal position for a given difference in load (w). 

The equation shows that for a given balance, [i. e. its weight 
(Wb) and the distance of its centre of gravity below the fulcrum 
(d) and the angle (a) which the arm makes with the horizontal 
are given] and a given load [2(W P + W) f-w] that 

1. The sensibility is proportional to the length of the arm. 
Hence, the beam should be as long as is consistent with 
strength and stiffness. 

2. The sensibility is proportional to the additional load (w). 

3. The sensibility varies inversely as the weight of the beam 
(Wb). Hence, the beam should be as light as possible. To 
fulfill this condition and (1), the beam is sometimes made of 
two hollow cones placed base to base or of a longitudinal sec- 
tion of such double cone with transverse braces. 

As the weight of the beam determines the amount of friction 
at the moving parts, the friction should be diminished as much 
as possible, and this is done by using steel knife edges and 
agate plates for bearings. 

4. The sensibility varies inversely as the distance of the cen- 
tre of gravity of the beam below the fulcrum (d). Hence, to 
increase the sensibility the centre of gravity should be below 
the fulcrum, but as near to it as possible. In the actual bal- 
ance the position of the centre of gravity can be varied 
slightly by screwing a small ball up or down. 

5. The sensibility increases with the load [2 (W p -f-W)-j-w], 
in the figure, where the points of suspension of the pans, A, 
and A 2 are above the horizontal line, but it can be proved that 



5' 

if A l and A^ are below the horizontal the sensibility decreases 
?vith the load. 

6. If W b d < [2(W p +W)+w] L sin. a, the denominator is 
minus; tan. 9 is minus; is more than 90 ; and if, at the same 
time w=o, tan = — o, that is, the beam will turn through 
180 and take the horizontal position. 

7. If W b d>[2(W p -hW)+w] Lsin.tf, the denominator is plus, 
tan. must be plus and if at the same time w — o, tan. == o. 

w L cos. a 



8. If W b d = [2 (W p + W) + w] L sin. a, tan. 9 = 







and if at the same time w = o, tan. 9 = , that is, the tangent 

o 

is hide terminate and the beam is in equilibrium in any po- 
sition. 

9. Now, it is desirable to make the sensibility independent of 
the load. The equation shows that this may be done by plac- 
ing the points A„ A , A 2 , in the same straight line, that is, in 
making the beam straight or # = o, provided that the load 
[2 (W p +W) + w] is not such as to make the denominator equal 
zero. For, if the denominator be zero, tan. — cc and = 90 , 
that is, the beam will take a vertical position. If these condi- 
tions are fulfilled, 

w L 
tan. — ^ry— r • 
W b d 

ga. Under the conditions given in 9, can only become 90 ° 
when w is infinity, for, then tan. = 00 and & = 90 , or 

gb. When d = o, that is, when the centre of gravity is at 
the fulcrum. Hence, the centre of gravity must not be at the 
fulcrum. 

If the arms are of equal length, if a = o and (W p + W)-f-w= 

Wp-j-W or w - o, tan. = o and the balance is true. 

A balance which does not fulfill these conditions is false. 

135. — A false balance may be detected by exchanging the 
places of the weights and the body. 



52 

136. — Prop. The true weight of a body, weighed upon a 
balance of unequal arms, is the geometrical mean of the tzvo appa- 
rent weights, when the body and the weights are made to exchange 
places. 

Place the body to be weighed in one pan and add standard 
weights to the other until balance is established. Then, ex- 
change the positions of the body and weights, establishing 
balance again. This is called the " Method of Double Weigh- 
ing." 

Let W = the true weight of the body. Let W r — the appa- 
rent weight of body when weighed in the right pan. Let Wi= 
the apparent weight of the body when weighed in the left pan. 
Let L r = the length of the right arm. Let Li = the length 
of the left arm. 

Then W-Li = W r -Iv r . 

W ■ I* = Wi . U. 

W 2 = WrWi. 

W = i/W r Wi. 

137. — Considerable degree of accuracy may be attained by 
"taring." By this method the body is balanced with sand 
or shot. The body is then removed and standard weights ad- 
ded until equilibrium is again established. The added stand- 
ard weights give the weight of the body. 

Let Wb = weight of the body ; W s = weight of the shot and 
W the standard weights. Put the body in the left-hand pan 
and the shot in the right-hand pan, then W b • Li = W s ■ L r . 

Now, remove the body and add standard weights, then 

W . Li = W 8 • Lr ; W b . Li = Ws . Li ; therefore, W b = W. 

138. — The steelyard is a lever of the first class, having 
unequal arms. The power, called the equipoise, is constant 
in value and moveable along the bar. The fulcrum is fixed. 

The steelyard is graduated in equal divisions, beginning at 
a point at which the equipoise must be placed in order to bal- 
ance the weight of the bar when its centre of gravity is in the 
long arm, or, at the fulcrum if the centre of gravity of the bar 
is at that point. The proof is as follows : 



53 

Let Fig. 91 represent a steelyard, A its fulcrum, g its cen- 
tre of gravity in the long arm, F 2 the weight, F x the power, 
W b the weight of the beam concentrated at g. Then by §50 

F 1 .AiA+W b .gA = F i: A 1 A. 

Place the equipoise F, at such a point, A'i, that equilibrium 
may be established. It will evidently be so situated, §93, 
that 

W b • gA = Fi • A/A, hence 

F, • AxA-f Fi • A\A = F 2 • A 2 A 

F, (A 1 A-hA/A) = F,-A,A. 

Fx • AxA'i = F 2 ■ A 2 A. 

Now, in this equation, F, and A 2 A are constant, from the 
construction of the steelyard : therefore, A t A/ varies as the 
weight F 2 . 

Cor. If the centre of gravity of the bar happens to be at 
the fulcrum. F, • A a A = F 2 ■ A 2 A, hence, A,A varies as F 2 . 

139. — The Danish Steelyard differs from the ordinary form 
in that the power, which is the weight of the bar, is constant 
and the fulcrum moveable. In Fig. 92 the beam is repre- 
sented with a ball at one end. A is the fulcrum, F 2 the weight. 
W b the weight of the bar is concentrated at g, the centre of 
gravity. By §50, when equilibrium is established, 

F 2 ■ A 2 A = W b • gA = W b (A 2 g-A 2 A)= W b • A 2 g-W b • A 2 A 

F 2 • A 2 A+W b ■ A 2 A=W b • A 2 g. 

A 2 A(F 2 -,-W b ) = W b -A 2 g. 
W b ;A 2 g 

AaA F 2 +W b 

This equation shows that the scale divisions are not equal ; 
for, making F 2 — W b , 2 W b , 3 W b , &c, and solving for A 2 A 
the results are different. If the values given F 2 form an arith- 
metical progression the distances from A 2 form an harmonic 
progression. 

140. — The Wheel and Axle. The wheel and axle con- 
sists of two cylinders, one of greater diameter than the other, 
with a common axis. The power is applied to the circum- 
ference of the larger cylinder, called the wheel, by means of 



54 

a rope when used for drawing water from a well, by a winch 
as in the windlass, by the hand, or a lever, as in the capstan, 
by a revolving floor as in the treadmill and by other methods. 
The weight is applied to the circumference of the smaller cyl- 
inder called the axle, by means of a rope or chain. The mode 
of connection is such that when the power moves in one di- 
rection the weight moves in the opposite direction. It is 
obvious (Fig. 94) that the wheel and axle may be regarded as 
compounded of a series of levers A,AA 2 , of the first class, 
passing through the centre A of the wheel and axle which is 
the common fulcrum. If the forces are inclined as in the di- 
agram, — the law of balance is the same as in §50, that is, F x ■ AP 2 
= F 2 • AP 2 , the power multiplied by its arm equals the weight 
multiplied by its arm. 

The same result may be obtained by resolving the power 
and the weight, each, into two components, one tangent to 
the circumference and the other along the elemental lever. 
The horizontal components will simply pull upon the axis, 
while the vertical components will produce equilibrium, that 
is, A^ • AiA = A 2 T 2 • A 2 A. 

On account of similarity of triangles, 

A a F x : A.T^rA.A : AP r 

.-. F, • AP 1 =A 1 T 1 ■ A,A. 

Also, F 2 • AP 2 =A 2 T 2 • A 2 A. 

Hence, as above, F, ■ AP, = F 2 • AP 2 . 

Cor. — The lines of action of the poiver and tveight are usually 
parallel. 

In this case (Figs. 93, 95) it follows that the arm of the power 
becomes the radius of the wheel and the arm of the weight the 
radius of the axle, hence, for equilibrium, the power multiplied 
by the radius of the wheel equals the weight multiplied by the radius 
of the axle. 

Cor. If the rope is so thick that the thickness caitnot be neg- 
lected, half the thickness must be added to either the radius of 
the wheel or the radius of the axle, or both, according as the 
power or weight, or both, are so applied. 



55 

141. — The Rope Machine consists of a rope passing over 
one or more pulleys. The parts of the rope are inclined to 
each other. 

In Fig. 96, let F x be the power fixed to a cord passing over 
the fixed pulley Pi and F 2 the weight attached to the movea- 
ble pulley P-2, the other end of the rope being fastened to the 
beam P n P 3 . L,et 9 be half the angle included by the two parts 
of the rope. The law of balance of the machine is obviously 
that of the point A. To balance A the forces F n F a , F s , must 
form the triangle F,AR, §85. Then, by §86, 

F x : F 2 ::sin. ARF, : sin. AFiR:: sin. RAF 3 : sin. F 1 AF 3 :: 

. 2 sin. 9 cos. 9 , ■ 

sin. & : sin. 2 9 = : 1 : cos. 0, that is, for balance, 

the power is to the weight as unity is to twice the cosine 

of the half angle formed by the two inclined parts of the rope. 

Cor. 1. Suppose the rope is straight; then 2^=180°, 

z= 9 o°, and F, : F 2 :: 1 : cos. 90 : 1 : 0, or, F, = - 2 = 00, that 

is, if a line has weight, to stretch it into a horizontal position 
will require an infinite force. Or, solving for F a1 F 2 — 0, that 
is, in order to stretch a line horizontally by a finite force, the line 
must be without weight. 

Cor. 2. Suppose 2 9 = 120 or f^ = 6o°, then F x : F 2 :: 
sin. 6o° : sin. 120 , that is, F, == F 2 , or three forces, making 
angles of 120 with each other, balance, when they are equal. 

Cor. 3. Suppose the weight is allowed to descend until 2 # 
= 0, then, Fi : F 2 :: 1 : 2 cos. 9 or F, : F 2 :: 1 : 2, that is, the 
rope machine becomes a moveable pulley with the two parts 
of the cord parallel. §142. 

142. — The Simple Pulley. — The pulley is a rope machine 
with parallel cords. The tension of a rope is transmitted over 
a smooth peg or rotating wheel, to diminish friction. The 
pulley may be either fixed or moveable, simple or compound. 
The simple pulley has only one rope. 

In the fixed pulley the law of balance is that the power 
equals the weight. 



56 

If a force is applied at F, (Fig. 97) by the u principle of 
transmissibility", §15, it is transmitted upward to A, around 
A ; therefore an equal force F 2 would have to be placed at the 
other end to balance it. §21. The object of the fixed pulley 
is simply to change the direction of the force, as in the well 
bucket, ships' rigging, derricks, fire escapes, &c. 

In the moveable PULLEY the law of balance is that the 
'power eqitals the weight divided by the number of parts of the 
string which sustain it. In the diagram (Fig. 98) an applica- 
tion of the "principle of transmissibility," shows that F x is 
transmitted to the strands (1) and (2) ; hence, the moveable 
pulley B is supported by a force 2F, ; hence, F 2 must equal 

F 2 
2F1, for balance, that is F 2 =2F 1 or F, : F 2 :: 1 : 2 or -^ 2 =- or 

r 1 1 

generally, if there are several strands, the power equals the 

weight divided by the number of strands which support it. 

In (Fig. 98^) Fj : F 2 :: 1 : 4, or if there are n pulleys F 1 : F 2 
:: 1 : n. The pull at A = 5F1 = F 2 +F> or if Wi is the whole 
weight of the pullies at the lower block F 2 -J- W 1 =4F 1 . 

In White's Tackle (Fig. 98^) F^F,:: 1:7. When the sheaves 
of a block have the same diameter, they turn with different 
angular velocities. In White's pulleys the radius is made 
proportional to the velocity, relatively to the block, of the ply 
of rope which it is to carry. 

In (Fig. 98*) 2F 1 + 2F T + 2F 1 +2F 1 = 8F t = F 2 . 

143. — The Inclined Plane is a plane inclined to the hor- 
izon, such as an inclined plank, inclined ways for raising 
materials to the tops of buildings, up-hill roads, railroads, &c. 

Let AB (Fig. 99) be the given plane,, and B a body on it. 
From g, the centre of gravity of the body draw a line repre- 
senting its weight F 2 . It is obvious, that to prevent the body 
from sliding down the plane some force F, must be applied. 
The resultant of F, and F 2 must be perpendicular to the 
plane, otherwise it may be resolved into two forces, one per- 
pendicular to the plane, the other parallel to it. The latter 
will produce motion down the plane, which is contrary to the 



57 

hypothesis of balance. Call the angle ABC, 0. Then, in the 
triangle F x gN 

F, : F 2 :: gF a : F,N :: sin. F x Ng : sin. F,gN :: sin. : sin. F x gN, 
that is, the power up the plane is to the zveighi as the sine of 
the angle of inclination of the plane, is to the sine of the angle 
which the power up the plane makes with a perpendicular to 
the plane. 

Cor. i. — Suppose V 1 is parallel to the plane, then FjgN^cp 

F 1 
andF^F^rsin. : sin. 90 :: AC: AB ::h:l or^ = — , that is, 

b 1 n 

the power parallel to the plane is to the weight as the height 

of the plane is to its length. 

Cor. 2. — Suppose F 1 is parallel to the base BC, then F x gN 
=90°— #. Fi : F 2 :: sin. : sin. (90 — #) :: sin. : cos. :: AC : 
BC : : h : b, that is, the power parallel to the base is to the weight 
as the height of the plane is to the base. 

144. — The Screw. If an inclined plane, the length of whose 
base is equal to the circumference of a cylinder be wrapped 
around the cylinder the plane becomes a screw. The height 
of the plane is the distance between the threads of the screw, 
and the base is the circumference of the screw. The screw 
works in a hollow cylinder with corresponding threads, called 
the nut. From this arrangement, it is obvious, that the power 
is applied parallel to the base of the plane and the law of 
balance is, §144, 

F, :F 2 ::h:b. 

F 27TT 

Fj : F 2 :: p : 2?rr, or —?==-—, in which r is the radius of 

the screw, that is, the power is to the weight or resistance as 
the distance between the threads — pitch — of the screw is to the 
circumference of the screw. 

The screw is used where great resistance is to be overcome. 
It is usually combined with the lever. A spiral staircase, a 
road winding around a mountain, are other illustrations. 

When the thread coils from the left hand over to the right, 
the screw has to be turned from left over to right to cause it 



5« 

to enter the nut ; if the thread coils from right over to left, it 
must be turned from right over to left. The former is called 
right-handed, the latter left-handed. 

145. — The Micrometer Screw. It is obvious, that, when 
the head of the screw (Fig. 101) is turned once, the screw ad- 
vances a distance equal to the distance between its threads ; 
suppose this distance is the 1-40 of an inch. Let the head be 
so large that it may be divided into 25 equal parts. If the 
head makes one rotation, the screw advances one-fortieth of 
an inch or, .025 inch ; if it is rotated only one twenty-fifth of 
a circumference, the screw will advance 1-25 of 1-40=1-1000 
=.001 inch. Suppose, for example, the head was rotated three 
turns and twenty divisions of the head, the advance will be .075 
+ .020=. 095 inch. By placing an object between the end 
of the screw and the stop (s) and turning the head until con- 
tact is made, very small dimensions may be measured ; hence 
the name micrometer. 

146. — The Isosceles Wedge consists of two equal inclined 
planes placed base to base. In the inclined plane the body is 
dragged over the plane ; in the wedge the plane is pushed 
under the weight or resistance. The length of the inclined 
plane is the /ace of the wedge ; the height of the plane is half 
of the back and the angle formed by the two faces is the edge. 
The remaining surfaces form the ends of the wedge. The 
power is applied perpendicularly to the back and the resist- 
ances perpendicularly to the faces, one half to each face. 

Let a b c (Fig. 102) represent a section, of an isoceles wedge, 
parallel to the ends. Let F 1 be the power applied normally 
to the back and ^F 2 the resistances similarly applied to the 
faces. Let 2 & be the angle of the wedge. . 

Produce the lines of action of these forces until they meet 
at A. Then by §86, 

Fi : K F 2 ::sin. F 2 AF/::sin. F 2 AF l ::sin. 2 Q = 2 sin. S cos. 
: cos. :: 2 sin. : 1. 

Fi : F 2 .: 2 sin. : 2 :: sin. 6> : 1 :: bd : be, that is, the power is 
to the resistance as half the back is to the length of the /ace. 



59 

The wedge is used where great force is to be exerted through 
a small distance, as in raising ships, buildings, splitting tim- 
ber and stone, extracting oils from seeds, &c. 

C. Compound Machines. 

147. — Definitions. — Compound Machines consist of two or 
more simple machines so connected that the weight or resist- 
ance of one becomes the power of the next, and so on. The 
law of balance can often be determined by writing down the 
law of balance of each simple machine in the form of a pro- 
portion, multiplying the proportions, term by term, and throw- 
ing out common factors. 

148. — The Compound Lever is represented in Fig. 103. 

For the first lever, Fi : F 3 :: A 3 A : A, A. 

For the second lever, F 3 : F 4 :: A 4 A' : A 3 A'. 

For the third lever, F 4 : F 2 :: A 5 A" : A 4 A", 

F, : F, :: A 3 A ■ A 4 A' • A 5 A" : A X A • A 3 A' • A 4 A", that is, the 
power is to the weight as the product of the arms on the side of the 
weight is to the product of the arms on the side of the pozver. 

Other forms of compound levers are shears, pliers, nut- 
crackers, tongs, the jaws of animals, &c. 

149. — The Knee-joint (Fig. 104) consists of three bars CA, 
AB, BD. CA is pivoted to an immovable support at C ; AC 
and AB, which are equal in length, are hinged at A. AB and 
BD are pivoted at B. The power is applied vertically down- 
ward at the hinge A, and BD moves horizontally through a 
guide. A body placed between D and B will be subjected to 
great pressure. The law of balance is, evidently, that of the 
point A. §86. 

Let BAC = 2-0. Then 

F 1 : F 3 :: sin. 2# : sin. & :: 2 cos. 0sin. # : sin. 9 :: 2 cos. : 1. 
But the resistance at D is applied horizontally, therefore, its 
component along AB only is efficient to produce balance, but 

sin. 

then, F x : . 2 :: 2 cos. 0:1. 
sin. 



6o 

F : F :: 2 cos. : sin. :: 2 : — — :: 2 : tan. 0, that is, ^ 
1 2 cos. # 

power is to the resistance as two is to the tangent of half the incli- 
nation of the two bars, or 

smce,tan.0=~\F 1 :F ! :: 2 :tan.0::2:|^::AP,:^BC, 

that is, the power is to the resistance as the height of the joint is 
to half the distance between the ends of the bars. 

Cor. From the vertical position of the bars to the horizon- 
tal the thrust increases until the efficiency becomes infinite, 
for, (i) let the bars be vertical, a condition expressed by 
making 2 # = 0, then Fi : F, :: 2 : 0, or F 2 = 0. 

Again, let the bars be horizontal . 2 = i8o°,0 — 90 , and 
F x : F 2 :: 2 : tan. 90 :: 2 : 00, that is, F a — 00. The enormous 
pressure exerted by this machine renders it very useful in 
printing, coining, and as a brake on locomotives. 

150. — The same result can be obtained by an application of 
the u principle of moments." Take C as the centre of mo- 
ments, then 

F.-CPH-F.-CP.^O; F 3 = 3- 

i sin. 

F 1 ■ CA sin. = F s ■ CA cos. ACP 3 = F 3 ■ CA sin. CAP 3 = 

F 3 • CA sin. 2 & = -r^4 CA sin. 2 0. 

sin. & 

F 1 .sin. 2 <9 = F 2 sin. 2 0. 

F, : F 2 :: 2 sin. cos. 8 : sin. 2 :: 2 cos. : sin. 0:2: tan. 0. 

151. — Roberval's Balance (Fig. 105) is in common use. It 
consists of a lever BC free to turn about the point A, having 
articulated at its extremities two vertical supports BE, CD 
upon which are placed two pans. The beam DE is supported 
at A' and at its ends the two vertical supports are suspended. 
In every position of DE, the supports BE and CD maintain a 
vertical position and BCDE remains a parallelogram. The 
position of the loads upon the pans does not affect the con- 
dition of equilibrium. Place equal weights Fi and F', unsym- 
metrically upon the pans. Apply at D and E two forces, each 



6i 

equal to F n but opposite to each other in direction. The stat- 
ical condition of the system will not be changed. F 1 and F/ 
are thus replaced by a single force F x at D, and a couple F n 
— Fi with arm Li, and a force F'j. at E and a couple F 1} — Ft 
with arm L, 3 . The effect of the couples is counteracted by 
the resistance of the points A' and A, and the result is the 
same as if the weights had been placed symmetrically in the 
centres of the pans. 

152 — The Compound Balance (Fig. 106) used for weighing 
heavy articles of merchandise is made in many forms. The 
ordinary hay-scales is represented in Fig. 106. ' Four equal 
levers of the second class, whose fulcra are A x , A 2 , A 3 , A 4 , di- 
verge from the four corners of an excavation. A platform not 
represented in the diagram has its weight transmitted to four 
knife-edges K a , K 2 , K„ K 4 . The power ends of the four 
levers rest upon a support, S, which rests upon a fifth lever of 
the second class, its fulcrum being at A 6 . The other end of 
the fifth lever is linked to a sixth lever, Ls, of the first class. 
The first four levers are evidently equivalent to a single lever. 
An application of the preceding principles will easily show 
the law of balance. 

153. — The Compound Wheel and Axle (Fig. 107) consists 
of a succession of wheels and axles. The power is usually ap- 
plied by means of a crank. The crank may be regarded as 
one spoke of a wheel with a handle placed at right angles to 
it. The weight is applied to the circumference of the axle. 
The wheels are called drivers or follozvers, according as they 
communicate or receive motion. The crank is a driven wheel 
or follower ; the final axle a driving wheel or driver. 

Let the radii of the axles be represented by r with sub- 
scripts, and the radii of the wheels by accents, then 



F, 


:F s ::r, 


:r'. 








P. 


:F 4 ::r 5 


:r" 








F 4 


••F,::r, 


. ^ttr 








F, 


:F 2 ::r,. 


V r s 


:r' 


■ r" 


r'" 



that is, the power is to the weight 
as the product of the radii of the drivers is to the product of the ra- 
dii of the followers. 



62 

154. — Wheel Work. The power may be transmitted from 
the circumference of the axle of one wheel and axle to the wheel 
of the next by means of a rope or belt as in Fig. 108. The 
law of balance is the same. §153. 

To prevent the slipping, arising from the use of belts, teeth 
may be placed upon the circumference of the wheels as in Fig. 
107. The mode of connection shows that the circumferences 
must move with the same velocity. Hence the number of 
revolutions of geared wheels varies inversely as the radii. Since 
the teeth must be the same size in geared wheels, the number 
of teeth becomes a measure of the circumference, and hence 
the number of revolutions varies inversely as the number of 
teeth. 

155. — The Differential Wheel and Axle — Double 
Axle, or Chinese Windlass, (Figs. 109, loga) is a wheel and 
axle, the axle having a thick and thin part. . The rope is so 
connected that when it winds up on the thick part of the axle, 
it unwinds on the thin part. The weight is attached to a 
moveable pulley which hangs from the rope. The power is 
applied through a crank, for one revolution of which the 
weight is raised a distance equal to the difference between the 
radii of the thick and thin parts of the axle. It is, therefore, 
equivalent to a simple wheel and axle, whose axle has a ra- 
dius equal to the difference between the thick and thin part. 

Fig. 1 oga represents a section of a double axle by a plane 
perpendicular to its axis. 

Let the radius of the wheel=R; the radius of the thick part of 
the axle = R t ; the radius of the thin part of the axle — r. 
Taking C as the centre of moments, 

F 1 .R4->iF 1 T=}^F 1 -R t . 

F 1 .R=^F 2 .R t -^F 9 .r=^F 2 (R t — r). 

F.iF.iij^Rt-rJiRiRf-riaR, 

F 2 R ■ . 

or — = that is, 

F, ^(Rt-r) 

156. — The power is to the weight as the difference between 
the radii of the thick and thin parts of the axle is to twice the 
radius of the wheel. 



157. — Weston's Pulley Block (Fig. no) is a modifica- 
tion of the differential pulley. In Fig. no the upper pulley 
is cast in one piece but has two radii. The power is trans- 
mitted by an endless chain which is prevented from slipping 
by projecting pins on the circumference of the axis. The 
power is applied to the circumference of the larger pulley. 
The direction of the chain is sufficiently clear in the figure. 
A comparison of this with Fig. 109 shows that a great saving 
is effected in the length of rope used for a given lift of the 
weight. 

158. — Definition. Compound Pulleys have more than one 
rope. 

159. — In Fig. in, beginning with the fixed pulley, 

F^F^il:!. 

F 2 :F 3 ::1 : 2. 

F 3 :F 4 ::1:2. 

F 4 :F 2 ::1:2. 

F, : F 2 :: 1 : 8 = 2 3 or 2 n if the number of cords is n. That 
is, the power is to the weight as one is to two raised to the power 
denoted by the number of moveable pulleys. 
F 



The pull on A is 

2 

The pull on A , is y 2 of 
The pull on A 2 is % of 



F F 

_2 ^_2 

2 "~ 22 

F F 

— = — 

2 2 2 3 



The total pull *= F [ - + ?.+ -,) =F t (1 — -s 

\2 2 2 / \ 2 

The whole weight on the support is F 2 ( 1 *) -f F,. 

160. — In Fig. 112, which is Fig. in inverted, the load on 
the lower pulley is 2F n on the middle one 4F n and on the up- 
per one 8F X . Hence, since the weight equals the sum of the 
loads, F 2 =F 1 + 2 F 1 + 4 F 1 =F, (1+2+20=-^ (2 3 — 1), or gener- 
ally if n = the number of pullies F 2 =F 1 (2 n — 1) ; that is, the 
power is to the weight as two raised to a power denoted by the num- 
ber of pullies, less one, is to one. 



6 4 

161. — In Fig. 113 the first set of pullies attached to the up- 
per horizontal support has its law of balance F 8 : F 2 : : 1 : 4 ; the 
second set called " luff tackles," F : : F 3 :: 1 : 3; hence, the law 
of balance for the whole system is F t : F 2 :: 1 : 12. 

162. — In Fig. 114 the law of balance is that F, : F 2 : 1 : 12. 

163. — In Fig. 115 Fj : F s ::l : 4. The pressure at A, is F, 
and at A 2 is 4F r 

164. — In Fig. 116 F 1 : F 2 :: 1 : 5 cos. 0. The pressure at A, 
is 2F T cos. # ; at A 2 is 4F a cos. #. 

165. — In Fig. 117 Fi : F 2 :: I : 4. 

166. — In the Pulley and Inclined Plane (Fig. 118) let 
F 2 be the weight of the body and F' 2 the component of the 
weight tending to cause it to slide down the inclined plane. 
Let F l be the power applied to the cord passing over the pul- 
ley, then 

F' 2 :F 2 ::l:h 

F, : F' 2 ::l : 2 

F 2 : F a :: 1 : 2h, that is, the power is the %v eight as the length of 
the plane is to twice its height. 

167. — The Screw and Lever. (Fig. 119). The screw is 
usually combined with the lever at right angles to the length 
of the screw. 

Let the radius of the circle described by the power = r' ; 
the radius of the head of the screw = r 3 . A is the fulcrum of 
the lever. 

F a : F 3 :: r, : r' :: 2ttt 1 : 2^r ; , 

F 3 : F 2 :: p : 2^r, 

F l : F 2 :: 2p?rr 1 : 2^ • 2^r r ;: p : 27rr', that is, the power is to 
the resistance as the pitch of the screw is to the circumference 
described by the power. 

168. — The Differential Screw (Fig. 120) consists of a 
cylinder AB upon which two threads T, and T 2 are cut of dif- 
ferent pitches. The screw has two nuts, one of which N x is 
fixed to a frame, the other N a is moveable in a groove. If the 
screw is turned through one revolution N 2 will be moved for- 



65 

ward a distance equal to the pitch of T, and backward a dis- 
tance equal to the pitch of T 2 or forward a distance equal to 
the difference between the two pitches. 

169. — Hunter's Screw (Fig. 121) is a modification of the 
differential screw and consists of a solid screw S x working in 
the interior of a hollow one S 2 . The lower end of the solid 
screw is attached to a platen M, through which the force is to 
be exerted. The solid screw works through its nut N, which 
is fixed in position by the frame work of the press. The 
pitches of the two screws are different, that of S 2 being great- 
est. When the large screw descends the smaller one rises in 
its interior. For one revolution of the head of the large screw 
the platen descends a distance equal to the difference between 
the pitches of the two screws. For example, let the pitch of 
the large screw be 1-20 inch and that of the small one 1-21 
inch ; for one revolution of the head of the large one the platen 
will descend 1-20 — 1-21 = 1-420 of an inch. It follows that 
this compound machine is equivalent to a simple screw, the 
pitch of which is 1-420 inch. 

170. — The Endless Screw is a combination of the wheel 
and axle with the screw. (Fig. 122.) It is used where a re- 
sistance is to be overcome through a great distance. 

Let r' = the radius of the crank ; x" = the radius of the 
wheel ; r 1 = radius of the axle ; p == pitch of the screw, then 

F, ; F 3 :: p : 2 77 r / 

F 3 : F 2 ::r, : r" 

. * . F, ; F 2 :: pr x : 271-rV, that is, the power is to the weight as 
the product of the pitch of the screw and the radius of the axle 
is to the circumference described by the crank multiplied by the 
radius of the wheel. 

171. — The Advantage or Efficiency of a Machine. The 

F 

ratio of the weight to the power, — 2 , is sometimes called the 

ri 

advantage or efficiency of a machine. An increase of this ex- 
pression may be effected by increasing F, when F, is constant 
or by diminishing F, when F 2 is constant ; a diminution, by 
diminishing F, when F, is constant or increasing F 1 when F 2 



66 

is constant. It is desirable to have F t very small, while F 2 is 

F 
large ; hence — ought to be as large as possible. 
Fi 

_ F 2 power-arm 

In the Lever, f^- = — r-z— ; hence, the advantage may 

' F, weight-arm & J 

be increased by lengthening the power-arm or diminishing 

the weight-arm. When the lever is of constant length this 

is effected by moving the fulcrum towards the weight. 

T A . . F 2 radius of the wheel , 

In the Wheel and Axle, — = — ,- ^ — , hence, 

b 1 radius of the axle 

the advantage is increased by increasing the radius of the 
wheel or diminishing the radius of the axle ; the former, be- 
yond a certain limit, would make the machine cumbersome ; 
the latter would diminish the strength of the axle. To obviate 
both difficulties the double axle, §155, has been constructed. 

F 1 
In the Fixed Pulley, - a — -, hence, its efficiency cannot 
Fi 1 

be increased. 

F n 
In the Moveable Pulley, ~ = - , hence, to increase the 

efficiency the number of parts of the sustaining cord must be 
increased. 

T ... T „ F 9 sine of the angle which the 

In the Inclined Plane, — - = 7 —. -. ~ - r — — -. — 

F, sine of the angle 01 mclina- 

power makes with a perpendicular to the plane , 

_ __ i — t _? hence, to m- 

tion of the plane, 

crease the efficiency the plane must be lowered or the sine of 
the angle the force makes with the normal to the plane in- 
creased ; but when the angle is 90 the sine is the greatest 
possible. 

F length 
If the power is parallel to the plane, ^ =«- . =- , hence, to 

increase the efficiency the plane must be lengthened or its 
height diminished. 

F hr-mf* 

If the power is parallel to the base, ^ 2 = -— -irp hence, to 



6; 

increase its efficiency the height must be diminished or the 
base lengthened. 

F 2 TCT 

In the Screw ^ 2 — , hence, to increase the mechanical 

1\ p 

advantage increase the circumference of the head or diminish 

the pitch. The former would make the machine unwieldy ; 

the latter would weaken the thread. The Differential Screw 

secures strength and efficiency without excessive size. §168. 

* ■ ' TTT F 2 length of the face , 

In the Isoscles Wedge, ^t = r — , , hence, m- 

' F 1 back 

creasing the length or diminishing the back will increase the 
advantage. 

Similar expressions may be obtained for compound ma- 
chines. 



68 



KINEMATICS 



172. — Definition. Kinematics treats of motion without ref- 
erence to its cause. It is a branch of geometry. 

173. — Distance, which is one dimension of space, is called 
space and is represented by s. It is always relative, that is, the 
distance of a point is always reckoned from some other point. 

Distance is a vector quantity ; hence, to define it, the mag- 
nitude, direction and sign must be stated. 

The magnitude of distance is called length. 

The unit of distance is the foot, which is one-third of the 
yard. 

The foot is divided into i-2 s , i-4 ths , i-8 ths , i-i6 ths , &c, or 

I-IO ths . 

"The British Standard Yard is the distance between the cen- 
tres of the transverse lines in the two gold plugs in the bronze 
bar deposited in the office of the Exchequer, (London) at the 
temperature, 62°F." 

174. — The French unit of length is the metre, which is the dis- 
tance between the ends of the platinum bar in the Archives, 
Paris, atO°C. The metre was intended to be the one ten-mil- 
lionth of the quarter meridian through Paris. But corrected 
measurements give 10,000,880 metres as more nearly correct 
for this distance. 
The metre is subdivided into i-io ths , i-ioo ths , i-iooo ths . 

175. — Distances of one point from another point. Im- 
agine one of two points fixed at A (Fig. 123). Let the position of 
the second point be at S. Then, the distance of the second 
point from the first is SA. Now, if we imagine the second 
point to be at a distance SiA above and to the left of A and 
also a distance S 2 A to the right of A, it is obvious that the 
real position of the second point is at S, the same as when it 
was said to be at a distance SA from the first point. 

Hence, if S,S is drawn equal and parallel to AS 2 , SA may 



6 9 

be regarded as the equivalent of the distances AS, and S,S or 
AS, 

It is evident that AS 2 could have been drawn first and then 
a distance equal and parallel to AS, from its extremity with 
the same result. 

In the above figure, complete the parallelogram AS,SS 2 ; it 
is obvious, that the distance of the second point above and to 
the left of A forms one side of the parallelogram ; the distance 
of the second point to the right of A is the adjacent side and 
the diagonal is the distance equivalent to As, and As 2 , hence, 
the following, 

176. — Prop. Parallelogram of Distance. If the two ad- 
jacent sides of a parallelogram represetit i?t magnitude {length) 
and direction two distances, of one point fro?n another, the di- 
agonal draw7i from the intersection of these two sides will rep- 
resent the resultant distance in magnitude and direction. 

Hence, for every proposition under the parallelogram of 
forces there is a corresponding one under distance. 

Hence, distances, like forces, may be compounded and re- 
solved. 

177.— rDistance and Several Points. Prop. If the distance 
of a given point from a fixed point is given, which fixed point 
is distant from another fixed point, the distance of the third 
point from the first is the resultant of the distance of third 
from second and the second from the first. 

Let a point (Fig. 124), which may be called the first point, 
be situated at S, distant SS, from a fixed point at S,, or second 
point. Let Si be distant S,A from a third point situated at 
A. Join A and S. AS will be the distance of the third point 
from the first. Complete the parallelogram AS,SS 2 and it 
will be seen that AS is the resultant of ASi and SS,. Hence, 
the truth of the proposition. 

178. — Motion. Definition. Motion is change of distance 
between two points. 

Motion is always relative, because distance is relative ; that 
is, the motion of a point is always in reference to some other 
point regarded as fixed. 



7° 

The distance between two points may change in magnitude 
(length), or in direction or in magnitude and direction. 

If the distance change only in length, the motion is recti- 
linear, if in direction and length, curvilinear, if in direction 
only, circular. 

178^. — Imagine one point to be fixed. 

If the distance AS between two points (Fig. 125) placed at 
A and S increase to ASi or decrease to AS 2 , the second point 
has moved relatively to the first the distance SS, or SS 2 or if 
(Fig. 126) the distance increase from AS 2 to AS 2 and the di- 
rection change from AS 2 to AS r 2 , . the second point has 
moved S 2 S' 2 . 

If AS 2 equals AS' 2 , S 2 S' 2 is the arc of circle. 

1 ySb. —Again, imagine both points to have motion. 

179. — 1. Let two points (Fig. 127) occupy the position Si and 
S 2 at the beginning of a certain time, and S', and S r 2 be their 
positions at the end of this time ; find the relative motion of 
the point at S 2 to S, or S, to S 2 ; that is, find a line which will 
represent the change in direction and length of the distance 
between Si and S 2 . 

Take a point at S (Fig. 128) regarded as fixed and draw SS 2 
and SS' 2 equal and parallel to SiS 2 , and S?\S t '. Join S 2 S' 2 , it 
will represent the required relative motion. 

180. — 2. Or regarding the point at Sj (Fig. 129) as fixed, 
draw S 2 S n and S 2 S / i equal and parallel to S/S./ and S,S X . 
SiS', will represent the required motion ; it is the same as 
S/S 2 (Fig. 128) but in the opposite direction. 

181. — 3. Or regarding the point (Fig. 130) as fixed at S/, 
draw S/S 2 equal and parallel to S^ but in the opposite direc- 
tion. Complete the parallelogram, the diagonal S/N will be 
the motion of the point at S 2 relatively to the point at S, ; 
hence, 

182. — Prop. I. " The Parallelogram of Relative Motions." 
If the two adjacent sides of a parallelogram represent in mag- 
niiude and direction, two motions, the diagonal drawn from 



7 1 

their intersection will represent the resulta?tt motion in magni- 
tude and direction, 

Hence, for every proposition under forces there is a corres- 
ponding one under relative motions ; hence relative motions 
may be compounded and resolved. 

183. — Prop. II. The relative motions of three points in a 
given time may be represented by the three sides of a triangle 
in magnitude and. direction. 

Take the three points (Fig. 131) 1, 2, 3. Suppose the one 
at O to be fixed. 

Then, S1S2 is the motion of one (2) of the points relatively to 
O and S'jS',,, the motion of the other (3) point relatively to O 
(1). Join S, and S/ and complete the parallelogram SjS^S/. 
S will be the position of the point (2) at S l5 but S r 2 is the real 
position of (2). Join S',S. In the triangle SS'gS',, SS/ is the 
motion of the third point relatively to the first, SS/ the mo- 
tion of the second relatively to the first. 

184. — Prop. III. The motions of any number of points may be 
represented by all the sides of a polygon, or the motion of the last rela- 
tively to the first will be the resultant of the motions relatively to 
those that precede. 

Take four points (Fig. 132) 1, 2, 3, 4 ; then their relative 
motions may be represented by the sides of a closed polygon 
or the motion of 4 relatively to 1 will be the resultant of the 
motion of 4 to 3, 3 to 2, 2 to 1. 

185. — Velocity. — Definitions. Since motion is change of 
distance and direction, it becomes necessary to determine the 
rate of change of distance and also of direction. For the rate 
of change of direction see §216. 

The rate of change of the distance between two points is 
velocity. It is measured by the distance described, ex- 
pressed in units of length, divided by the time during which 
the distance is described, expressed in units of time ; that is, 
s 

v = f 

The elements of velocity are : magnitude, sometimes called 
"speed 1 ' ; sign ; direction. 



72 

The direction of the velocity of a point is the direction of its 
path if straight, or if the point move in a curve the direc- 
tion of the tangent to the curve at the point under considera- 
tion; for example, let a point move in the curve AS, (Fig. 133). 
The direction of the velocity at A is the tangent AT. 

The unit of distance is the foot. § 1 73. The unit of time is 
the second, which is the 1-86. 400 th of a mean solar day, hence, 
the unit of velocity is one foot per second which may be called a 
vel. 

Velocity may be uniform or varied. 

186. — In Uniform Velocity equal distances are described 
in equal intervals of time. Thus if a body move over two 
feet in each second the velocity is uniform. 

1 87. — Prob. Find the distance described by a body moving tvith 
uniform velocity in a straight line. 

Let s = the required distance. 

Let v = its velocity, i. e. the number of feet described in 
one second. 

Let T = the time of the motion in seconds. 

Then s = Tv, that is, in Uniform Velocity, the distance de- 
scribed equals the time multiplied by the velocity. This is true 
whether the path be straight or curved. 

Cor. 1. The time taken by a body to move over a given 
distance with a given velocity, when the velocity is uniform 

equals the space divided by the velocity. T=-. 

v 

Cor. 2. The velocity acquired by a body which has moved 

over a given distance in a given time in uniform motion equals 

s 
the distance divided by the time, v = — 

If two uniform velocities are compared it is obvious that 

Cor. 3. The distances vary as the product of the times and 
velocities, that is, s,: s a :: tjV,: t 3 v 2 . 

Cor. 4. If the velocities are equal, the distances vary as the 
times, that is, s,: s. :: t :t 2 . 



73 

Cor. 5. If the times are equal, the distances vary as the ve- 
locities, that is, s x : s 2 :: v, : v 2 . 

Cor. 6. If the spaces are equal, the velocities vary inversely 
as the Amu, that is, v^ v 2 :: t a : t r 

188. — Let two distances (Fig. 134) Asi and As 2 , due to the 
velocities v, and v 2 be given. Complete the parallelogram 
As^; then As is (§176) the resultant distance of As, and As 2 . 

Draw Avj and Av 2 coincident with or parallel to As, and 
As 2 . Then, if the parallelogram Av,vv 2 be completed Av will 
be its diagonal and will represent the resultant of Vi and v 2 . 
The two triangles Avv 2 and Ass 2 have their sides Av 2 and As 2 , 
and vv 2 and ss 2 parallel, and if t is the time during which the 

motions take place, since Av x = — x -\ Av 2 = — - . * . Av, : Av 2 :: 

As,: As 2 or vv 2 : Av 2 :: ss 2 : As 2 , that is, the parallel sides are pro- 
portional. Hence, Av the diagonal of Av 1 vv 2 is the result- 
ant of v, and v 2 , just as As is the resultant of As, and As 2 , hence, 
the 

189. — Prop. "The Parallelogram of Velocities." If the 
two adjacent sides of a parallelogram represent in magnitude and 
direction two velocities impressed upon a body, the diagonal drawn 
from their intersection will represent the resultant velocity in mag- 
nitude and direction. 

Hence, for every proposition under forces there is a corres- 
ponding one under velocity. 

190. — Acceleration. Definitions. Velocity is not always 
constant. Varied velocity may be accelerated or retarded. 

The direction of motion may be nnform or varied. 

In varied velocity the distances described in equal intervals 
of time are unequal. Thus, if a body move over 2, 5, 8, 12, 
14, &c. feet in successive seconds, the velocity is varied. If 
the distarices described in successive seconds increase the ve- 
locity is said to be accelerated ; if they decreased is retarded. 

If the increments of velocity in equal times are equal the 
velocity is uniformly accelerated ; if the decrements are equal, 
uniformly retarded. Thus, if the numbers 1, 3, 5, 7, 9, 1 1, 



74 

&c, represent the spaces described in successive seconds, the 
velocity is uniformly accelerated ; if 1 1, 9, 7, 5, 3, 1, represent 
the spaces described in successive seconds the velocity is uni- 
formly retarded. 

A body falling in vacuo has uniformly accelerated velocity ; a 
body projected upward in vacuo has uniformly retarded velocity. 
In all cases of varied velocity ^ the velocity at any instant," means 
the velocity the body would have if it continued to move uni- 
formly with the velocity at that instant, for the time under consid- 
eration. Thus, a train of cars moving with the speed of sixty 
miles per hour, does not necessarily move sixty miles in the 
hour, but would, if it continued to move uniformly for the 
hour with the velocity at the instant observed. 

The rate of change of linear velocity is called linear accel- 
eration, and is measured by the velocity described, expressed 
in units of velocity, divided by the time during which the 
velocity is described, expressed in units of time, that is, 

v 
a = _. 

T 

The elements of acceleration are, magnitude, sometimes 
called "quickening''; sign ; direction. 

The direction of acceleration is the same as that of velocity. 

Retardation is negative acceleration. 

The unit of velocity is one foot per second ; the unit of accelera- 
tion is one foot per second per second, and may be called a eel. 

191. — Prop. "The Parallelogram of Accelerations." If 
the two adjacent sides of a parallelogram represent in magnitude and. 
direction, two accelerations impressed upon a body, the diagonal 
drawn from their intersection will represent the resultant in magnitude 
and direction. This proposition follows immediately from the 
" Parallelogram of Velocities" (§189) since v = aT. 

Hence, for every proposition under force we have a corres- 
ponding one under acceleration. Hence, the composition and 
resolution of accelerations. 

192. — Uniformly Accelerated Velocity. Prop. I. The fi- 
nal velocity of a body having uniformly accelerated -velocity equals 
the initial velocity plus the acceleration multiplied by the time, that is, 
v 2 = Vl -f-aT. 



75 

Take a series of numbers, i, 3, 5, 7, 9, 1 1, 13, &c. If each 
number represent a velocity at the end of each second, the 
series will represent uniformly accelerated velocity in which 
2 is the acceleration ; since 2 is the increase of velocity for 
each second. The velocity at any second can be easily deter- 
mined; e. g. find the velocity at the end of the fifth second. 

The increase of velocity for 1 second = 2 ; the increase of 
velocity for 5 seconds = 10. But the initial velocity =1 ; . * . 
the final velocity v 2 = 1 + 10 = 1 1. 

193. — To make the case general let a body start with an 
initial velocity v, ; let the acceleration be a, and the time of 
the motion T ; let v 2 = the final velocity, then, 

v 2 = v, + aT ; for, by definition, the velocity is increased 
a for each second; for T seconds the increase will be aT; but 
the initial velocity is v 1? therefore the final velocity equals the 
initial velocity plus the acceleration multiplied by the time. 

1 94. — Prop. II. The space described in uniformly accelerated 
velocity equals tlie initial velocity multiplied by the time plus 
one-half the acceleration into the square of the time, that is 
s = v,T -f- }^aT 2 . 

195. — First method. Let The the whole time of the motion; 
let a be the acceleration and \\ the initial velocity. Imagine 
T to be divided into a great number (n) of small intervals (t) 
of time all equal to each other ; then T = nt. 

The times are 

t 2t 3t (n— i)t nt 

The velocities at the be^innin^s of these intervals are 
v 1 Vj+at v, + 2at Vj+3at Vj+in— i)at 

The velocities at the ends of these intervals are 
Vj-f-at Vj+2at vi+3at v x -f(n — i)at Vj-fnat 

The distances passed over with these velocities are 
v x t v,t+at2 v 1 t+2at 2 v x t + 3at 2 v,t+(n— i)at 2 

v>t+at 2 v 1 t + 2at* v t t+3at 2 v 1 t+(n— i)at 3 v^+nat 2 
The sums of these distances are 

n Vl t+at 2 (i+2-r-3 n— 1)^= Vl T+— f i—-- 

2 \ n 



7 6 

n Vl t + at (i-h2 + 3- • • • n) -vj-h— 2 f i +1 

2 \ n 

The distance actually described must be somewhere be- 
tween these two sums. By making n very large, - will be- 

aT 2 

come very small, and the two sums will differ from vT -f 

2 

aT 2 

by less than any assignable quantity, hence s = v,T -f- 

196. — Another method. The average velocity is the veloc- 
ity a body moving with varied velocity must have, in order to 
move uniformly over the same distance. It is obvious that 
the space described can be determined by multiplying the 
time by the average velocity. 

T , 1 -4. : v-hv- v^Vj-r-aT a T 

ihe average velocity is — -— — — v : + — 

22 2 

The space described s= v, -f- — T= v : T -+- Y / 2 aT 2 . 

197. — Cor. 1. The square of the final velocity, in uniformly 
accelerated velocity, equals the square of the initial velocity 
plus twice the acceleration multiplied by the distance de- 
scribed, i. e., v 2 2 = Vj 2 + 2as. 

For v 2 = v, 4- aT. §192. 

s = viT+J^aT 2 . §194. 

. * . v,» = v 1 2 + 2v 1 aT -f- a 2 T 2 = v, 2 + 2a (v/T + % aT 2 ). 

. * . v 2 2 = v, 2 -f 2as. 

198. — Cor. 2. If the body start from rest a fact expressed by 
making v, = o. 

The final velocity equals the acceleration multiplied by the 
time, v 2 = aT. 

The distance described equals one-half the acceleration mul- 
tiplied by the square of the time, s — ^aT 2 . 

If at the same time T= 1, s= >^a or 2s = a, i. e. the accel- 
eration equals twice the space passed over in the first second. 

The distance equals half the product of the time and final 
velocity, s 2 — j£Tv 3 ; and 



77 

The sqtiare of the final velocity equals twice the prodtict of 
the acceleration and the space described, or v,, 2 = 2as. 

199. — Cor. 3. If the velocity is uniformly retarded, that is if 
a is negative, these equations become 
v 2 = v, — aT. 

s = v,t-^aT 2 . . • 

v 2 2 = v^ 2 — 2as. 
If the body start from rest v 2 =— aT; s— — y 2 aT 2 ; v 2 == — 2as. 

200. — A good illustration of uniformly acceleration velocity 
is furnished by a body falling freely under the action of grav- 
ity. General Morin's apparatus (Fig. 135) consists of a verti- 
cal cylinder of wood made to rotate by means of a weight. A 
strip of paper divided into rectangles is fastened upon the cyl- 
inder. The horizontal divisions will represent equal portions 
of time if the cylinder rotates uniformly. This is effected by 
attaching a fan-brake to the top of the cylinder. A small 
body, in guides, to which a pencil is attached, may be released 
when the cylinder is moving uniformly. As the body falls a 
mark is made upon the paper. When the paper is developed 
it presents the appearance of Fig. 1350. 

The diagram shows that after the expiration of the first in- 
terval of time; the body has fallen vertically downward one 
space; at the end of two time divisions it has fallen four spa- 
ces; at the end of three intervals of time nine spaces, &c, or 
the result may be written : 

Times, 12345 &c. 

Spaces fallen, 1 4 9 16 25 &c. 

This experiment shows that the velocity is uniformly accel- 
erated for the spaces described are proportional to the squares 
of the times from rest. But this is (§198) the law of uniformly 
accelerated velocity. 

201. — It is evident that if the space described in one second 
were given, all distances for given times could be calculated. 
Now, experiments made directly upon falling bodies and with 
the pendulum establish the fact that a body in this latitude 
falls about 16. 1 feet per second ; hence the spaces described in 
1 second, 2 seconds, 3 seconds, 4 seconds, &c, are : 



78 

Times in seconds, o i sec. 2 sec. 3 sec. 4 sec. &c. 
Spaces described, o 16 64 144 256 &c. 

202. — The distances described in equal successive portions 
of times, are as the numbers 1,3, 5, 7, &c. 
Time in seconds, 1st sec. 2d sec. 3d sec. 4th sec. &c. 

Distances, 16 48 80 112 

These numbers are as 1 3 5 7 &c. 

203. — Since the distances 16, 48, 80, 112, &c, aredescribed 
each in one second, they represent velocities; the acceleration 
may be obtained by taking their common difference, which 

is 32. 

Hence, a falling body moves with uniformly accelerated 
velocity ; it falls through a distance of 16. 1 feet in the first 
second and has an acceleration at the end of the second which 
would carry it over 32.2 feet. §198. On account of the im- 
portance of this constant (32.2) it is represented by a special 
letter, g, which is read " the acceleration due to gravity." g 
varies at different elevations and at different parts of the 
earth's surface. 

204. — The equations under uniformly accelerated velocity 
become applicable to a falling body when a is made g and s 
becomes h. 

v f = vi4-^T §192. 

h = Vl T + ^<7T* §194. 

v., 2 = v, 2 + 2^h ; (§197) or, 

if the body start from rest, 

v 2 =^T 

h-^T 2 =^Tv 2 

v 2 

v 2 2 = 2gh or v 2 = i/ 2 ^h, and h=-— 

d 

205. — The expression V 2gh is called the velocity dice to the 

height h. 

v 2 
206. — The expression — is called the height due to the ve~ 

j 
locity v. 



79 

207. — Velocity of a Body Projected Downward in Vacuo. 

The equations of §204 will solve this case. 

2 o8. — If a Body is Projected Upward in Vacuo the equa- 
tions of §204 become 

v 2 = Vl -^T 
h = v 1 T-y 2 ffT* 

V 2— V 2—2ffh. 

The first equation shows that with a given initial velocity, 
as T diminishes, v 2 increases ; 

1. If gT = v. or T =— '-, v = o ; that is, 

9 
If the time of motion equals the initial velocity divided by 
the acceleration due to gravity, the body will rise to a height 
where its velocity is zero, that is, to the greatest height ; 

2. If T>— \ the final velocity is negative, that is, the body 

begins to return to earth. 

3. The height to which a body will rise when projected ver- 
tically upward equals the difference of the squares of the initial 
and final velocities divided by twice the acceleration due to 
gravity ; for, finding the value of T in the first equation and 
substituting it in the second 

2 9 

4. The greatest height to which a body may be projected ver- 
tically upward with a given velocity equals the square of the 
initial velocity divided by twice the acceleration due to gravity, 
for when the greatest height is reached the final velocity, 
v 2 — o ; substituting this value in the preceding equation, 

h — —j which is the expression formerly, §206, called the 

height due to the velocity v x . 

5. For any two times at equal intervals before and after reach- 
ing the greatest height the velocities are equal; for, find the ve- 



So 



locity for the time T =— l +T,; it willbe v 9 =v 1 — I- 1 — T \ n 

9 \g I 

= — gT x ; again find the velocity for the time T = — — T o , it 

will be v 2 = gT,. 

6. If a body be projected vertically upward, the velocity at any 
point in the ascent is the same as the velocity at the same point in 
the descent. For if the values v 2 = gT, and v 2 = — gT 1 be sub- 

2 2 

stituted in the equation, h — Vl 2 , the result is the same, 
namely, h = v t 9 — ^ 2 T, 2 . 

' If the point of projection of the body is taken as the point 
in the ascent, the body will rise to a height from which it must 
fall to acquire the velocity of projection. 

j. The times of ascent and descent are equal. 

209. — Velocity of a body down an Inclined Plane.— 

Suppose a body (Fig. 136) to slide or roll down an inclined 
plane without friction. Let g be the acceleration due to 
gravity when falling freely. Resolve this acceleration into 
two components ; one, g Y parallel to the plane, the other per- 
pendicular to it. The former will be the acceleration of the 
sliding or rolling body on the plane. Now, if is the angle 
of the olane, h its height and 1 its length, . 
• 'a h 

Hence, the velocity down an inclined plane is the same in 
kind as that of a body falling freely, the acceleration being 
diminished in the ratio of g : g sin. 0. It is for this reason that 
Galileo could use bodies rolling down inclined planes to de- 
termine the laws of free falling bodies. 

The general equations of uniformly accelerated velocity will 
apply if g sin. 6 be substituted for a. That is, 

v 2 = vi + g- - -T. §192. 
s= Vl T + ^.j.T 2 . §194. 



If the body start from rest at A as origin a condition expressed 
by making v, = o, 

v,=^-T. 

Several interesting results may be obtained by solving these 
equations. 

Prop. I. The time required to move from the top of the "plane to its 
bottom varies directly as its length and inversely as the square root 
of its height, for making s = 1 and substituting in the equation 

for the space, 1 = % g ■ - T 2 and solving for T, T = 1 ^2 

1 i/^h 

Hence, it follows that 

Prop. II. The times down planes having the same height but 
different lengths are proportional to their lengths. (Fig. 136^). 
Also, 

Prop. III. The velocity a body has acquired at the bottom of the plane 
equals the square root of twice the product of the acceleration due to 

gravity and the height of the plane, i. e., v a =V3 gh.. This is 
the quantity which was called " the velocity due to the height 
h." For, substituting the value of T into the general equa- 
tion of the velocity, v 2 = V 2gh ; hence the fact may be ex- 
pressed as follows : The velocity acquired by a body moving down 
an inclined plane is the same as if it fell freely through a vertical 
distance equal to the height of the plane. 

The special case of the Chords of a Circle which end at 
the extremities of a vertical diameter. 

Prop. IV. The velocity acquired by a body rolling down any chord 
through the extremity of the vertical diameter of a circle varies as the 
length of the chord, for, in Fig. 137 let I, be the length of any 
chord andh, its height, let r be the radius of the circle, then, the 

velocity at the bottom of the plane is v 2 —1/3. ghi 2 9 ' ^ 

V 2r 



82 



— 1 l 2 ff 

~ *i "y — , but the radical is constant, hence the velocity va- 
ries as the length of the plane. Also, 

Prop. V. The time of descent is the same for all such chords 
and is equal to the time of falling through a distance equal to the 



diameter-, for, T = l, -yj -^ = 1, -y g .\* = yj-- • 2r, that is, the 

^ 2r 

time is proportional to 2r or the diameter, for all chords. 

210. — Projectiles in a Vacuum. A body projected ob- 
liquely upward and abandoned to the attraction of gravitation 
is called a projectile. 

The curved path it describes is called its trajectory. 

The point of starting is called the point of projection. 

The distance on a horizontal plane from the point of projec- 
tion to the point where it again strikes the plane is called its 
range. 

The time of flight is the time consumed in describing the 
path from the point of projection to the point where it once 
more meets the horizontal plane. 

The angle of elevation or projection is the angle the initial 
direction of motion makes with the horizontal plane. 

2 1 1 . — Prop. The equation of the path of a projectile is y— x tan . 
ft — — x 2 , in which x and y are the co-ordinates of any 

2V 1 2 COS. 2 # 

point in the path, v^ the velocity of projection, ft the angle of 
projection and g the acceleration due to gravity. 

In Fig. 138 let O, the origin of the co-ordinate axes Ox, 
Oy, be the point of projection, the curve line be the trajectory, 
OT = v a the velocity of projection, the direction of which is 
tangent to the path at the beginning of motion and TOx = ft, 
the angle of elevation. Resolve OT into two component ve- 
locities, one along Ox, which may be called v x , the other v y> 
along Oy. Then v x — v, cos. 6 ; v y = v, sin. ft. 



83 

Now, it is obvious that v y is velocity due to a body project- 
ed vertically upward, and v x a uniform velocity. Since, the 
equation of a curve is an expression of the relation between 
the co-ordinates of any point in a curve it will be necessary to 
find the relation between x and y of any point in this curve, 
that is, of vertical and horizontal distances described with ve- 
locities v x and v y , hence 

y = v,sin. &T-y 2 gT 2 . 
x = v, cos. 0T. 

Eliminating T from these equations and solving for y the 
desired equation is obtained. 

i. Since this equation is of the second degree between two 
variables the path must be a conic section. 

2. If the equation be put in the form for completing the 
square, and the square completed, it will take this form : 
v 2 2 cos. sin. 0\ 2 2Vj 2 cos. 2 I v x 2 sin. 2 



■- 



x— -i =_^ y- 

1 g ' g l 2 q 

m . . . . . v 2 cos.0sin. 

Take the point whose co-ordinates arex— x,4- 

9 

v 2 sin 2 
and y = y 1 -f -J !_ and substitute their values in this equa- 

2 g 

2v* cos. 2 
tion, then x^ = — y, ; but this is the equation of a 

is 

parabola (Fig. 139) with the origin at its vertex, O, and of the 
form Xj 2 = — 4ay lt The directrix is + a above Ox n the focus 
— a below. The length of the latus rectum GH = 4a. In 

. , 2v 2 cos. 2 # v 2 eos. 2 # 
this particular case 4a — • a = — 

g *g 

3. If the body strike the plane at H, y, at that point equals 
zero and the equation of the path becomes 

= xtan. 0. 

2V, 2 cos. 2 

2v 2 sin. cos. v 2 -sin. 20 ,, . . j7 7 

x = — l - — -i , that is, the range equals 

9 ff 



8 4 

the product of the square of the initial velocity and the sine of 
twice the angle of elevation divided by the acceleration due to 
gravity. It follows that the expression will be greatest when 
9 is 45 ; hence 

4. The horizontal range is the greatest possible when the an- 
gle of elevation is ^5°, when it is equal to twice the height due 
to velocity v x . If 9 — 90 the projectile will ascend verti- 
cally. 

5. Also, for angles equally above and below ^j° the range 
must be the same. 

It is evident that the trajectories are not the same, for if 9 
and 90— # be substituted in the equation of the trajectory y 
will have different values. 

6. The time required to reach any point in the path equals the abscissa 
of the point divided by the product of the velocity of projection and 

the cosine of the angle of elevation, x— v a cos. 9 t ; t= 

V! cos. 9 

7. The time of flight along a horizontal plane equals twice the 
velocity of projection and the sine of the angle of elevation divided 
by the acceleration due to gravity, for 

M x 2v. 2 sin. B cos. 2v, sin. 9 , ~ . 

f = — — 1 — — l , hence Too sin. 

v, cos. 9 g v, cos. 9 g 

8. The time of flight is greatest when 9 is greatest, for a 
given range and initial velocity. 

212. — The Motion of a point is known in every respect 
when (1) its velocity and (2) direction are given. 

213. — Uniform Motion. Motion is uniform when both the 

velocity and direction are uniform, that is when equal distances 

s 
are passed over in equal times — or when v = — is constant, 

and the path is straight This subject was discussed in §187, 

214. — Varied Motion is due to change in the velocity or 
the direction or change in both the velocity and the direction. 



85 

Uniformly accelerated velocity in a straight line, or varied 
motion due to change of velocity without change of direction 
was treated in §192, &c. 

215. — Uniform Velocity in a Circle. A body moving with 
uniform velocity in a circle has its motion varied from change 
of direction of its path. 

216. — Prop. I. The rate of change of direction (or acceleration 
towards the centre or deviation from a tangent) equals the square 
of the uniform velocity in the circle divided by the radius of the 
circle. 

For, let a particle move in the circle (Fig. 140) whose 
centre is C, with iCniform velocity v. Let A, be its posi- 
tion at the beginning of an interval of time t. Draw 
the tangent A X T, to represent in direction and magnitude 
the uniform velocity of the point. Let A 2 be the posi- 
tion of the point at the end of the time t and let the tan- 
gent A 4 T 9 = AjTj be the direction and magnitude of the ve- 
locity. Draw the radii CA n CA 2 . The arc A,aA 9 == vt. The 

chord A,A 2 = vt. -^-^L_. Draw A.T/ equal and parallel to 
arc 

A^. Complete the parallelogram. A 2 T 2 is the resultant of 
A 2 T/ = A 1 T 1 and A 2 T 3 = T/T 2 , which is the change of direc- 
tion of the motion in the time t. Now, CA,A 2 and A 2 T/T 2 
are similar, therefore 

T/T.rAiA, :: A 2 T 2 : CA,. 

chord 

. ' . The acceleration T 'T, = ] a - AlL? = vt arc ' V = 

CA, 

1 r 

v 2 t chord XT . r , . . , , . , ,. . 

— • JNlow, 11 t is imagined to decrease without limit, 

r arc. 

, . c chord 
the ratio 01 approaches unity and the rate of deviation 

becomes ^. 
r 

217 — Prop. II. Hence, if a particle move in a circle its 
velocity is made up of two components, one tangential, the other 
normal.. 



86 

The tangential acceleration a t is represented by AT in Fig. 
141, the acceleration towards the centre of curvature a c by 

AN. 

218. — If the particle move in any other (Fig. 142) curve, 
and if at any instant in its motion, three points at equal dis- 
tances from each other but indefinitely near be taken the mo- 
tion in the curve may be regarded as the same as the motion 
in a circle coinciding to an indefinite approximation with the 
curve at that place. Hence, using the "osculating" circle, 
calling its radius, />, the radius of ^curvature of the curve the 
foregoing result may be applied to all curves in which a par- 
ticle is moving ; that is, the normal acceleration = — 

P 

219. — Curvature. Change of direction may be measured 
by the curvature of a curve. 

The curvature is the rate of angular change of direction of 
the tangent per unit length of arc. In Fig. 143 let the origi- 
nal position of the point be A 1 and the direction of the velocity 
A^; after the arc Au\ 2 has been described A 2 is the position 
of the particle and A 2 T 2 the direction of its velocity. Produce 
A 2 T 2 back until it cuts AiTj. The tangent has evidently 
moved through the angle T X ET 2 == #, while the particle de- 

T FT 
scribed A.A„. Now,— 1 2 is called the curvature. 

220. — Prop. The curvature of a circle varies inversely as 
the radius of the circle. 

For CA1T1 = 90 ; A X CA 2 = 90 - A/TiA, ; A 1 ET 1 = 90 — 
AiTjA,. 

.; . A X CA 2 — AsETj or the angle at the centre equals the an- 
gular change of direction of the tangent. Let p = the radius 
of the circle ; then, since = A 2 ET X and since 

V : AxA,:: 1 : p. 

p6 = A l A, 
& 1 

arc p ' 



§7 

v 2 v 2 l 

Cor. i. The acceleration — may be written — = v 2 c or the 

P . P 

product of the square of the velocity and the curvature. 

221. — Angular Velocity. In §185 linear velocity was de- 
fined to be the ratio between the distance described and the 

s 
time during which the motion took place or v = - . Now, a 

point which moves in reference to a fixed point may be re- 
garded as moving in a path around the fixed point and the ve- 
locity may still be measured by the length of the path divided 
by the time — or it may be more conveniently measured by the 
angular velocity. For example, in a circle (Fig. 144), the 

linear velocity of the point A in the circular path is v = -, 
but if B be the angle subtended by the arc s and r the radius 
of the circle, v = — , that is, 

The linear velocity equals the product of the radius and the 
angle in circular measure divided by the time. 

If a point at the extremity of unit radius be taken, its linear 

velocity will be to = = , which expression is called angular 

velocity. 

222. — Prop. I. The linear velocity of a particle moving in 
any circle equals its angitlar velocity multiplied by the radius 

of the path, v = — r = tax. 

223. — Cor. The velocity of a point A around O is the same as 
that o/O around a parallel axis passing through A, for this ve- 
locity depends only upon to, r and the direction of the axis 
which are the same in both cases. 



224. — Prop. II. Rate of deviation or normal acceleration 
2 ' in terms of angular velocity equals the square of the angu- 



@ 



lar velocity multiplied by the radius of the circle which the par- 
ticle describes, or a n = — = — - = C ° 2 P- 

P 



88 

225. — An angle may be measured in "radians" instead of in 
degrees. Draw a circle with radius unity. The length of 
the circumference equal 2~r = 2X3. 14159 = 6.28318. Now, 
any angle may be measured by the ratio of the length of arc 
it subtends to the radius. Select as unit angle, the angle 
whose are equals radius unity and call it one "radian." Any 
angle then may be measured in terms of this unit. 

In degree measure, the radian = ^ =57° -29578 =57° 

17' 44". 86. The circumference equals zk radians ; a right- 
angle = - radians. 

s 2 

226. — Since angular velocity 'is a directed quantity, to repre- 
sent it geometrically, (1) draw a line perpendicular to the 
plane of the angular velocity — this is the axis of angular ve- 
locity ; (2) cut off a part of the line proportional to the an- 
gular velocity, and (3) look along the line so that the angular 
velocity may appear right-handed. Compare with couples §30. 

Hence, for every proposition under couples there is a cor- 
responding one under angular velocity. 

Hence, angular velocities can be compounded and resolved. 

227. — Prop. III. The Parallelogram of Angular Velocities. 
If the two adjacent sides of a parallelogram represent in magnitude, 
position of axis and direction two angular velocities, the diagonal 
drawn from their intersection will represent the magnitude, position of 
the (instantaneous) axis and the direction of their resultant. 

228. — Prop. IV. The resultant angular velocity of a point about 
two points in the same plane is an angular velocity about a point so 
situated, on the line joining the two points as to divide the distance be- 
tween them inversely as the magnitudes of the angular velocities about 
those points, and its magnitude is equal to the algebraic sum of the 
angular velocities. 

Let a point (Fig. 145) have angular velocities co 1 about O, 
and co 2 about O a . Let xy be the plane containing the points 
about which the motion is reckoned. 



8 9 

Draw the lines 0,P, and 2 P 2 so as to represent the angu- 
lar velocities. Since these lines are both perpendicular to 
the plane xy, they are parallel and the resultant will be OP 
= co — co J -\- w 2 in magnitude and so situated that w 2 : ft* 2 :: O a O 
: 0,0 and w 1 :co:: 2 : 2 0,. Compare with §24. 

Cor. 1. If ftj, is right-handed and ft* 2 left-handed, O, (Fig. 
146) will be on Oj0 2 produced. 

Cor. 2. If ft*, and co 2 are opposite in sense and equal (Fig. 
147), that is, ft»,=ft> 2 , they will forma u couple of rotations," 
O will be at infinity and co l -\-(—(o 2 ) — °. The velocity of the 
point then is a linear velocity equal to ft*, . 0,0 3 , perpendicular 
to Oi0 2 , and in the plane which contains it. 

If all the points in a flat plate have equal and opposite an- 
gular velocities, all points in the plate will have equal and 
parallel velocities, that is, the motion will be a translation. 
§231. 

228#. — Prop.V. The resultant of a linear velocity v = ft*, ■ 0,0 2 
and an angular velocity ft*,— co in the same plane is a linear velocity 
equal, parallel to and in the same direction as the original linear ve- 
locity but removed from it a distance equal to the angular velocity 
divided by the linear velocity. The proof is the same as that of 
§68. 

The same truth may be stated as follows : An angular veloc- 
ity around any point is equivalent to an equal angular velocity in the 
same sense around any other point combined with a linear velocity. 

Let ft\ (Fig. 148) be an angular velocity around Oi. At 
any other point 2 apply two equal and opposite angular ve- 
locities. The resultant velocity will not be affected, for the 
last two balance each other. 

Now, w, and — ft*/ are equivalent to a linear velocity (Cor. 
2) co ■ o i o 2 and the angular velocity co 1 remains. 

229. — Angular Acceleration. When angular velocity va- 
ries, its rate of change is called angular acceleration and is 
measured by the angular velocity divided by the time, that 



is ft* 



T 



go 

230. — Angular acceleration, like angular velocity, is a di- 
rected quantity; hence, for every proposition under angular 
velocity there is a corresponding proposition under angular 
acceleration. 

231.— Motions of Rigid Bodies. Definition. Translation. 

If all the particles of a rigid body (§7) or system of particles 
move in parallel directions and with equal velocities, or if 
lines joining pairs of points move parallel to themselves with 
equal velocities, the body or system is said to have a motion 
of translation. If these lines are not in the same plane, then 
every plane in the body, in any position of the body, must be 
parallel to its original position before motion. 

If three points in a rigid body, not in the same line, move in 
parallel directions, with equal velocities, all the points in the 
body will move in parallel directions and with equal velocities. 
These motions, like all others, are relative to a fixed point. 
The paths described may have any form provided that the 
above conditions are fulfilled. 

The forward or backward motion of the piston of a steam 
engine illustrates translation. 

232. — Rotation. If lines in a rigid body or system of par- 
ticles change their direction the body has a motion of rotation. 
Suppose first that the moving system is in a plane. 

Take a line A, A/ (Fig. 149). If it take the position A 2 A/, 
A,A/ and the system connected with it have obviously rota- 
ted around the point O. 

In Fig. 150, let A^/ be the original position of a line and 
A 2 A/ the final position. Join the points A, and A 2 and bisect 
the line A,A 2 at P. Similarly bisect A/A/ at P'. At P and 
P r erect perpendiculars; they will intersect at O. If A, A/ 
be rotated around O it will take the position A 2 A/, since 
OA,A/ and OA 2 A/ are equal in every respect. 

The axis of rotation is a line in the rigid body whose direc- 
tion is not changed by the motion. ' 

The plane of rotation is any plane perpendicular to the axis. 



9i 

Angular motion is the angle between perpendiculars to the 
axis at the beginning and the end of motion. 

The speed of a rotating piece is frequently expressed in 
turns and fractions of turns per minute, hence it becomes 
necessary to express angular velocity in terms of revolutions. 
Let co be the angular velocity and R the number of revolutions. 
One revolution equals 360 or 2rc. Hence co = 2n R. 

233. — Prop. I. Every possible motion of a rigid body can be re- 
solved into a motion of translation and a motion of rotation. 

Let the motion be in the same or parallel planes. 

In Fig. 151 take a line AiA/ and after a certain time let it 
take the position A 2 A/. If the point A 1 be given a motion 
A,A„ every point in the body will be moved parallel to itself 
and over the same distance and A l will be superposed upon 
A Q — that is, the body has a translation ; if A X A/ be now rota- 
ted through the angle which A, A/, A 2 A/ make with each 
other, A,A/, and therefore the whole body will take the posi- 
tion A 2 A/, that is, there is a rotation. 

Or, if A 2 ' move to A/ and then rotation be made through the 
same angle in the same sense, the result will be the same as 
before. 

234. — Prop. II. A translation may be regarded as equivalent 
to two equal and opposite rotations in the same plane. §228. Let 
a line A, A, in (Fig. 152) in a rigid body be given a transla- 
tion to A/A/. It is obvious that the same effect could be 
produced by rotating A X A, around its right end and then ro- 
tating about its left end through the same angle but in oppo- 
site directions. 

235. — Rotations and translations, though treated separately 
are one and the same, for a translation may be regarded as a 
rotation about an axis at an infinite distance. For, by the 
previous statement, §234, a translation is equivalent to two 
equal and opposite parallel rotations, but by §228 these are 
equivalent to a rotation about a point at an infinite distance. 

236. — Let the motion be in any direction. The definitions of 



9 2 

translation and rotation enable us to determine the motion of 
a rigid body from the motions of three of its points ; for if the 
velocities of three points have components parallel and equal 
to each other there is translation ; if there are components 
perpendicular to intersecting planes passing through the 
points there will be rotation. The line of intersection of the 
planes will be the axis. 

237. — Let BB (Fig. 153) be a rigid body and O a fixed point 
relatively to which the body moves. Let A,, A 2 , and A 3 be 
any three points in the body. Let A{V^ A,V 2 , and A 3 V 3 rep- 
resent the velocities of these points in magnitude and direc- 
tion. Construct a new figure. (Fig. 154.) Take any point 
o and draw lines oa n oa 2 , oa 3 , equal and parallel to v„ 
v 2 , v 8 . Now resolve oa n oa 2 and oa 3 each into two compo- 
nents, so that there shall be six components, the first three of 
which are equal and parallel. This may be done by drawing 
the plane a t a 2 a 3 , and resolving oa n oa 2 , oa 3 along and per- 
pendicular to the plane. oP is a common component of the 
velocity of three points ; therefore there is translation. 

Again from the extremities of the lines representing v,, v 2T 
v 3 draw lines representing the velocities perpendicular to 
the plane z y a 2 a 3 , but opposite in direction and complete 
the parallelogram. The other components of v„ v 2 and v $ 
will be the velocities parallel to the plane a^a.,. Now 
pass planes through A 2 and A 2 perpendicular to these com- 
ponents, they will intersect in OO.CX, parallel to OP. A 
plane drawn through A 3 will intersect in the same line. Now 
1 A 1 , 2 A 2 being perpendiculars not changed by the motion, 
OO.O, is the axis. The three components A, V/, A 2 V/, A 3 V/ 
are proportional each to the perpendicular distances (^A,, 
2 A 2 , 3 A 3 because A/V 7 / = co - OA ; A 2 V/ = <» ■ 2 A 2 ; 

A V ' 
A 3 V/ = co ■ 3 A 3 ; hence> the angular velocity co = r ^- L = 

A.V/ A,V/ 

2 A 2 3 A,' 

Hence, the motion is like that of a screw in its nut — that is, a ro- 
tation with a translation parallel to the axis of the rotation. 



93 

238. — Cor. If the translation equals zero there is rotation. 
If the angular velocity of rotation equals zero there is trans- 
lation. 

If the plane of rotation is given the velocities of two points 
only are necessary to determine the axis of rotation. 

239. — Prob. When the points in the same plane have par- 
allel motions oblique to the plane. 

240. — Prob. When three points in the same plane move 
perpendicularly to the plane. 

Let Aj, A 2 and A 3 (Fig. 155) be the three points; (1) if their ve- 
locities are equal the motion is a translation, (2) if their veloci- 
ties are unequal the motion is a rotation about the axis which is 
the intersection of the plane of the three points with the plane 
drawn through the extremities v x , v 2 , v 2 of the three lines 
which represent the velocities. The angular velocity is found 
as above in the general problem. 

Sometimes the general method fails, at other times a more 
simple method is available. 

241. — Prob. Let the motions of the points of the body be 
all parallel to one plane. 

Two points only are needed. Draw two planes (Fig. 156) 
perpendicular to the directions of the velocities through A 1 
and A 2 . If the planes cut each other the motion is rotation, 
if they are parallel the motion is a translation. 

241*2. — Prob. When three points, not in the same plane 
have parallel motions. 

242. — The axis of rotation may change its direction in space, 
but for any instant it may be regarded as fixed. It is then 
called an instantaneous axis. 

Illustrations of the Instantaneous Axis. 

243. — Suppose a body has a rotation around an axis and the 
axis a motion of translation in the same plane. A wagon 
wheel rolling over a horizontal plane is such a body. 

In Fig. 1 57 let a body rotate about an axis whose projection 



94 

is O, left handed, with angular velocity «; also let the axis O 
have a velocity of translation Ov at right angles to its own di- 
rection. 

Let fall a perpendicular OP upon the tangent TT, which is 
parallel to Ov. Let the point P be so selected that its distance 
from the axis equals the linear velocity of translation divided 

by the angular velocity of rotation or so that OP = — or v = 

CO 

co • OP. It is clear that the point P will be moved towards the 
left with a linear velocity, v, due to the translation and 
towards the right with a linear velocity co ■ OP (§222) but this 
last, from the mode of selection of the point, equals v. Hence 
the two velocities will balance and P will be at rest for the 
instant in question. It is apparent that every point, in a line 
perpendicular to the plane of rotation passing through P is 
equally an instantaneous point, hence such a line will be an 
instantaneous axis. If any point A, in the body be taken, it 
is obvious that at the instant considered^ linear velocity will 
be A t T, = co . A, P. Its direction will be tangent to a circu- 
lar arc of radius AiP or perpendicular to A,P. 

A,P is called the instantaneous radius-vector. 

Hence, any point, in a body, having a rotation and a transla- 
tion in the same plane at right angles to the axis of rotation, has a 
resultant velocity round an instantaneous axis, situated parallel to 
the original axis, at a distance from it, equal to the velocity of 
translation divided by the angular velocity, and the magnitude of 
the velocity equals the angular velocity of the body multiplied by the 
instantaneous radius vector of the point. 

244. — Mechanical Generation of the Path. A cylinder 
rolled upon a plane passing through P and parallel to Ov will 
obviously give the path of the moving point. 

If the point considered is O (Fig. 157) its path will be a 
straight line parallel to the plane. 

If the point is outside the circumference of the cylinder 
(Fig. 158) a curtate trochoids produced. 

If the point is in the circumference of the rolling cylinder 
(Fig. 159) a cycloid will be generated. 



95 

If the tracing point is within the circumference of the roll- 
ing cylinder (Fig. 160) a prolate trochoid is generated. 

All these curves are included under the general name of 
trochoids. 

245.— A Body may be Subjected to Parallel Rotations in 
the same Plane. 

Case i. — The angular velocities are both right-handed. 

Let a body (Fig. 161) be rotated around the point 1 with 
angular velocity co i and let O, be rotated around 2 with an- 
gular velocity o> 2 . By Prop. IV, §228, the resultant angular ve- 
locity will be around an axis whose projection is P, so situated 
that it will be parallel to the two axes Oi and O, and divide 
the distance between them inversely as their angular veloci- 
ties or so that co :w i: w 2 ::OA: O a P: OiPoro*, ■ 1 P=<V ° 2 P - 
Its magnitude equals their algebraic sum or co=w 1 -{- co 2 . Now, 
the point P or a perpendicular to the plane of rotation at P is the 
instantaneous axis ; for, its linear velocity to the right due 
to the angular velocity around 2 is w 2 • Q P, and the velocity 
to the left due to the angular velocity O t is co i . OiP, but 
these two expressions are equal ; therefore the axis is at rest. 

If any point A : in the rotating body be taken its velocity is 
perpendicular to the instantaneous radius vector and equal to 
the angular velocity multiplied by the length of the instanta- 
neous radius vector. 

Case II. The angular velocities are opposite in direction 
and co x is greater. (Fig. 162.) 

By reasoning similar to that in Case r, P is the instanta- 
neous axis ; the resultant angular velocity is co = co x — co 2 ; 
and co : co x \ co 2 :: 2 O a : 2 P : 0,P and the linear velocity of any 
point equals the resultant angular velocity multiplied by its 
instantaneous radius vector. 

Case 3. Let co i and «„ be opposite in direction but w 2 >w 1( 
(Fig. 163.) The resultant angular velocity is co — co 2 — co x . 

246. — Mechanical Generation of the Paths. If a cylinder 
be imagined to roll upon a fixed cylinder the resultant path 
of any point in the rigid body may be exhibited. 



9 6 

In Case i a convex cylinder rolls upon a convex cylinder. 
Let the radius of the small rolling cylinder be to the radius of 
the fixed cylinder as i : 3. If the tracing point is at the centre 
of the rolling cylinder the path will be a circle. (Fig. 164.) If 
the tracing point is between the centre and circumference of 
the rolling cylinder, (Fig. 164) will also exhibit the path. 

If the tracing point is in the circumference of the rolling 
cylinder, Fig. f 65 gives the path. 

If the tracing point is beyond the centre of the rolling cyl- 
inder, Fig. 166 exhibits the path. 

Other Illustrations of Case i. — A convex cylinder 
rolls on a convex cylinder of the same diameter. If the tra- 
cing point is in the circumference of the rolling cylinder a 
cardioid is generated. Fig. 167. 

2. A convex cylinder rolls on a convex cylinder of twice 
the radius. Fig. 168 shows the nature of the epicycloid. In 
these two cases one-half is the ratio of the diameters. 

247. — Case 2. A small convex cylinder rolls in a large con- 
cave cylinder. 

Figs. 169, 170, 171, show the various paths. 

Fig. 169 shows the prolate epitrochoid and Fig. 171 the cur- 
tate epitrochoid. 

Other Illustrations of Case 2. 

1. Let the diameter of the rolling cylinder equal the radius 
of the fixed cylinder. 

a. Each point in the circumference of the rolling cylinder 
describes a simple harmonic motion along the diameter of the 
fixed cylinder. (Fig. 172.) 

b. Each point in the axis of the rolling cylinder describes 
a circle of the same radius as the rolling cylinder. (Fig. 172.) 

c. Each point on or beyond the rolling cylinder describes 
an ellipse with its centre at the centre of the fixed cylinder. 
(Figs. 172, 173.) 

248. — In Case 3 a larger concave cylinder rolls on a smaller 
convex cylinder. 



97 

All the preceding curves are called " epitrochoids." An 
epicycloid is an epitrochoid described when the tracing point 
is in the circumference of the rolling cylinder. 

249. — When two equal, opposite and parallel rotations are 
combined, the resultant angular velocity is zero and the motion 
is a translation with a velocity equal to the common angular 
velocity multiplied by the distance between the parallel axes. 
§228. Cor. 2. 

Fig. 174 represents a connecting rod 0,0/, attached to two 
equal cranks all moving in the same plane. If the plane con- 
taining 2 0, has a right-handed rotation with angular velocity 
w 2 around 2 and the connecting rod a left-handed rotation 
around O, with angular velocity co 1 equal but opposite to co^ 
it is seen that every point in or attached to the rod has a trans- 
lation in a circular path whose radius equals 2 0,. 

It is obvious that this case cannot be represented by a roll- 
ing motion. 

250. — Rotations about Inclined Axes. A rigid body may 
have simultaneous rotations around axes inclined to each 
other. 

Let a body (Fig. 175) rotate, right-handed, looking from O, 
to O around the axis 0,0 with angular velocity w, and let the 
plane 0,00 2 rotate around 2 right-handed looking from 2 
to O with the angular velocity a> t . 

Now, by Prop. Ill, §227, if a part of 00, is cut off propor- 
tional to the angular velocity co i and from 00 2 a part propor- 
tional to co 2 and the parallelogram OAPB completed the diag- 
onal OP will represent the resultant angular velocity. 

OP will also represent the direction of the instantaneous axis. 
O must be a point on this axis, because it is a fixed point. 
Any other point in it must be determined by the condition 
that the linear velocity due to the rotation around 0,0 must 
be equal and opposite to the linear velocity due to the rotation 
around 2 or that w 1 • PP X = co 2 • PP 2 or that PP, : PP 2 : : ^ 2 : co x . 
But PP, :PP 2 ::sin. P,OP:sin. P 2 OP, and sin. P,OP : sin. P 2 OP 
:: (o % : w,. Now, OP can only fulfill this condition when it is 



9 8 

the diagonal of a parallelogram the two adjacent sides of which 
represent the angular velocities, the positions of the axes and 
the directions of rotation. §226. 

251. — Mechanical Generation of the Paths. If a cone roll 
upon a fixed cone any point in the rolling cone will describe 
the required path. The curves are spherical epitrochoids. 

252. — Prop. The motion of a rigid body may be always re- 
garded at each instant as a translation along an axis passing 
through the centre of mass of the body combined with a rotation 
about the same axis. 

The proof obviously follows from the preceding proposi- 
tions. §§237, 189, 64. 

253. — Rotation of a rigid body is known when its angular 
velocity and the direction of its axis are given. 

254. — Uniform Rotation. Rotation is uniform when both 
the angular velocity and the direction of the axis remain un- 
changed during the motion. 

255. — Varied Rotation is due to change in the angular ve- 
locity or the direction of the axis or change in both the angu- 
lar velocity and the direction of the axis. 



99 



KINETICS 



256. — Definitions. Kinetics is the science which treats of 
the relation between motion and the force which produces it. 

S11. 

In Kinetics inertias are compared. 

257. — The idea of mass or the quantity of matter in a body 
as distinct from weight or the pull of gravity upon a body is 
obtained from the following considerations : 

Let several bodies of exactly the same size and shape but 
of different materials, be placed upon a horizontal plate of 
glass so smooth that when motion is produced there shall be 
no friction ; imagine also that the air offers no resistance to 
the motion of the bodies. Let the bodies be of lead, clay and 
cork. It is obvious that the attraction of gravitation is bal- 
anced by the resistance of the glass and therefore need not be 
considered. 

Now, it is a matter of every day experience, 

(1) that these bodies will remain at rest so long as they are 
not acted upon by an external force. Again, 

(2) if the bodies are subjected to the action of an external 
force — after the action ceases they will continue to move uni- 
formly and in a straight line until compelled to change their 
velocity or direction by the application of another force — or a 
body under the action of no force is either at rest or moves 
with uniform velocity and in a straight line. These two 
facts are included in Newton's law of inertia or 

258. — First Law of Motion, Every body perseveres in its 
state of resting or of moving uniformly in a straight line, 
except so far as it is compelled to change its state by impressed 
{external) forces. 

259. — A ball lying upon a whirling table, the inertia ap- 
paratus, a man at rest upon horseback, the starting of teams 



IOO 

and trains are familiar illustrations of the tendency of bodies 
to remain at rest when an attempt is made to move them. 

Bodies in motion tend to remain in motion, with the same 
velocity as is shown when a rider is thrown over the horse's 
head when the horse suddenly stops, or when passengers are 
thrown forward when trains are suddenly stopped, or when a 
person leaps from a rapidly moving train or carriage ; the 
direction of motion is also preserved as is shown by the tan- 
gential course taken by mud or water when thrown from a 
wheel and by the difficulty one feels in changing the direc- 
tion of motion when moving rapidly. 

To change the direction of motion alone a force must be 
applied at right angles to the original direction of motion ; 
for, if applied at any other angle it may be resolved into two 
components, one in the original direction, which will accel- 
erate or retard the velocity, and the other at right angles, 
which will change the direction. 

260. — Again, imagine the same constant force to be applied 
horizontally, for & given time, separately to each one of the 
bodies : 

(1). Each one will move. §258. 

(2). Each one will move in the direction of the force. 

(3). Each will acquire a certain velocity which is different 
for the different bodies. The cork will show the greatest ve- 
locity, the clay less, and the lead the least. Hence, different 
accelerations have been produced by the same force, a fact 
which may be expressed by saying that a force zvhich will pro- 
duce a given acceleration is proportional to the mass. 

(4). Again, if the force is doubled each body will acquire 
an acceleration double that due to the original force ; if treb- 
led, three times that due to the original force, etc. ; a truth 
which may be expressed by saying Wiethe force is proportional 
to the acceleration it produces. 

(5). If a single body is taken — lead, for example — and an 
acceleration a is produced by a given force in a given time, sup- 
pose one second — in two seconds double the acceleration will 



IOI 

result — a fact expressed by saying that a given force acting 
upon a given body produces an acceleration which is propor- 
tional to the time. 

That time is an element in the transmission of force is illus- 
trated by great weights being moved by a comparatively small 
force if time is used ; locomotives start trains of separate cars 
which they would be unable to move if continuous ; a rapidly 
moving pistol ball makes a smooth round hole because time 
is not given to communicate its action except to the particles 
in front ; near shots are less destructive than far, and the 
entrance of a musket ball in the flesh is distinguishable from 
the exit. 

These facts Newton has expressed as follows in his 

261. — Second Law of Motion. Change of mass-velocity {mo- 
mentum) is proportional to the impulse which 'produces it and is in 
the same direction. 

262. — Definitions. Mass-velocity (mv), momentum or quan- 
tity of motion is proportional to the product of the mass and its 
velocity : if the unit of momentum is properly selected mass- 
velocity equals the product of the mass and velocity. 

Mass-acceleratio7t (ma), change of mass-velocity, rate of 
change of momentum, acceleration of momentum or inertia is 
the product of the mass and the change of velocity or accel- 
eration. 

If ma is called inertia, m or mass may be called the co-effi- 
cient of inertia. 

The unbalanced force (F,) multiplied by the time (TJ dur- 
ing which it acts is called an impulse (F.Ti). 

263.— Mathematical Expression of the Second Law of 
Motion. 

Let F, be a force, which acting on a mass m, produces an 
acceleration of a, eels in time T x . 

Let F 2 be another force which acting on a mass m 2 produces 
an acceleration of a, eels in a time T 2 . Then the law is pro- 
properly expressed as follows: F X T X : F 2 T 2 :: m^: m 2 a 2 . 



102 

If units of force, mass and acceleration be properly selected 
the proportion may be reduced to an equation. For example, 
if we take as unit force, a force which acting on unit mass will 
produce unit acceleration in unit time the proportion becomes 

F.T,: 1 x lnm^: l X l, or, 
FiTi=m 1 a 1 . 

The expression may be further simplified by making T\=l, 
then 

Fi=m 1 a 1> that is, a force may be measured by the acceler- 
ation which it will produce when acting on a given mass for 
one second. 

264. — Definitions. Units. The absolute unit of force is that 
force which acting on unit mass will produce unit acceleration 
in unit time. 

The British Unit of Mass is the mass of a certain piece of metal 
kept in the Exchequer Office in L,ondon. This unit of mass 
is called one pound avoirdupois. 

The French Unit of Mass is the gramme. 

The poundal is the unit of force when the pound is taken as 
unit of mass and one foot per second per second the unit of acceler- 
ation. 

The dyne is that force which acting on the gramme will pro- 
duce in one second an acceleration of one centimeter per second per 
second. This is the force unit of the centimetre — gramme 
— second or C. G. S. system. 

Application of the Second Law to Bodies Moving* in 
Curved Paths. 

265. — In Kinematics, §217, it was shown that when a point 

moves in a curve its motion is due to two accelerations ; one 

in the direction of a tangent to the curve at the point under 

consideration called the tangential acceleration a t ; the other 

normal to the path called rate of deviation or normal acceleration, 

v 2 
a n , the value of which is ---. 



io 3 

If the body have mass m, its tangential mass -acceleration is the 
mass multiplied by the acceleration (ma). 

The normal ma,ss-acceleration is the mass multiplied by the 

v 2 
normal acceleration or m - . This is sometimes called devia- 
te 

tion of momentum. 

Now, the second law (§261) of the relation between force 
and motion simply states that the tangential mass-acceleration 
equals the impulse which produces it and the normal mass- 
acceleration equals the impulse which produces it. 

For example (Fig. 176) let a mass m describe the arc of a 
circle because acted upon by the force F. The acceleration 

v 2 
along the arc has two components a t and — . The correspond- 

v 2 
ing mass-accelerations are ma and m - . Resolve F into two 

components, one in the direction of the tangent, the other* 
along the normal, then 

F t T = ma 

v 2 
F a T = tn-, 

P 

or, if T is made equal to unity 

F t = ma 

P 
266. — Mass in Terms of Weight. The first of these equa- 
tions gives a definition of mass. For, if the force acting is 
gravity (W), a becomes the acceleration due to gravity or g 

W 

and W = m<7, that is, m = — , or the mass of a body equals 

if 

its weight divided by the acceleration due to gravity. §9. 

Cor. 1. Any force can then be expressed as a weight multi- 
plied by the ratio between its acceleration and that due to 

gravity : for F = ma ; F — — . a = W • -• 

g 9 



104 

Cor. 2. Any acceleration can be expressed as the accelera- 
tion due to gravity multiplied by the ratio between the force 
producing it and the weight of the moved \ for 

F 
W — — F 

F = ma = — -a ; a = W -^r. — . 

ff 
267. — The Third Law of Motion. To an action there is always 
an equal and apposite reaction, or the actions of two bodies upon each 
other are always equal and opposite in directions. 

The following illustrations given by Newton include all 
cases : 

If one presses a stone with a finger, the finger is pressed by 
a stone. This fact was familiarized in Statics. 

If a horse draws a stone fastened to a rope, the horse is 
drawn backwards equally towards the stone. 

If one body impinges upon another and changes the motion 
of the second body, its own motion suffers an equal change 
in the opposite direction. 

The ordinary illustrations given in text books, of a boat- 
man pressing with a pole against the shore, the use of a cane, 
rowing, flying, swimming, the action and reaction between a 
gun and the shot, between a magnet and its armature, a 
broadside and the ship from which it is fired, the collision of 
bodies, such as balls, trains of cars, etc., can be arranged under 
the above examples. 

Illustrations of the Second Law of Motion. 
268. — Gravity or the Attraction of Gravitation. §108. 
Definitions. — Gravity is that force which acts between all 
masses of matter in the universe. 

This attraction between masses is mutual and equal. See 
Newton's third law, §267. For example, take two bodies, one 
containing one molecule, the other one thousand molecules. 
Now if we admit that the one molecule attracts each of the 
thousand molecules of the second body as much as each of the 
thousand molecules attracts the one, then the thousand mole- 



io 5 

cules exert a thousand units of attraction on the one and the 
one a thousand units on the thousand. 

The direction of gravity is towards the centre of the earth, 
for all observations on free falling bodies show that they fall 
in that direction. 

The point of application of the force of gravity in a body 
is its centre of gravity. 

The magnitude of the force of gravity depends on 
(i) the mass of the body, and 

(2) on the distance of the body from the centre of the earth 
— according to the following relations: 

1. The force of gravity is proportional to the mass of the body 
upon which it acts. 

Experimental Proof. If two bodies, having different masses 
m n m 2 , be dropped from the same height, at the same instant, 
they will, in vacuo, both reach the earth at the same instant. 
Now, Newton's second law, §261, shows that force is propor- 
tional to the change of momentum it will produce in a given 
body in a given time, or F x : F 2 :: m^: m 3 m 2 . 

But in this experiment F, is the force of gravity acting on 
the first mass m n F 2 the force of gravity acting on the sec- 
ond mass m 2 , a, is the acceleration due to F x and a 2 the accel- 
eration due to F 2 . The experiment shows that a x and a 2 are 
equal ; therefore, F gl : F g2 :: m,: m 2 , that is 

The force of gravity is proportional to the mass upon ivhich it acts. 

2. The force of gravity varies inversely as the square of the distance 
from the centre of the earth. The proof of this proposition can be 
found in Astronomy. 

In Fig. 177, let EE represent the earth and C its centre. 
Suppose a body of mass m 1 to be placed at A x on the surface 
and another body m 2 at A 2 above the surface. The law states 
that the force of gravity acting on the mass nil at A, is to the 
force of gravity acting on m 2 at A 2 as the square of the dis- 
tance of A 2 from the earth's centre is to the square of the dis- 
tance of A, from the earth's centre, or in mathematical symbols 



io6 

> a 1 1 

F gi : F g2 :: A 2 C : A,C or F gl : F„ :: = = Combining this 

AjC ' A 2 C 

proportion with the preceding, the whole law of gravity may 

be expressed as follows : 

p gl : F g 2 :: 71 • T* that is 

», » 2 

269. — The attraction of gravitation is proportional to the 
mass of the body divided by the square of its distance from 
the centre of the earth. 

This proposition is exceedingly important. The law of 
electrical and magnetic attractions and repulsions is the same. 

The following remarkable consequences follow from this 
law: 

270. — Cor. 1. The attraction of a uniform sphere upon a 
particle exterior to it is the same as if all its matter were 
concentrated at its centre. The geometrical proof is given in 
Newton's Principia, or see 

27 1. — Cor. 2. A particle situated within a hollow sphere is equally 
attracted in every direction and therefore will remain at rest in any part 
of the void. 

Let EE (Fig. 178) be a spherical shell and P any point within 
it. Through the point draw two lines intercepting very small 
arcs on the circumference. The angles a,Pb, and a 2 Pb 2 are 
similar and a,^: a 2 b 2 :: a r F: a 2 P. 

The very small areas generated by allowing the lines to re- 
volve around an axis through P and in the plane of the paper 
will be proportional to the squares of the lines, that is, the 
elemental area at a 1 = m 1 : the elemental area at a 2 = m 2 :: 

a7? : aj>*. 

But the law of attraction is 

Fg,:Fgi:: a^'a7 

F gl .F g2 .. =- -. =-5 ..1.1- 
a,P a 2 P 

therefore F ff .=F ffa . 



Ail other very small surfaces may be taken oppositely in the 
same way and the same result will follow. Hence, the result- 
ant attraction due to the whole spherical shell upon the en- 
closed particle will be zero. 

272. — Cor. 3. In a solid sphere of uniform density gravity va- 
ries directly as the distance from the centre. 

Let EE (Fig. 179) be a solid sphere with its centre at C and 
imagine a particle situated at A,. The sphere may be re- 
garded as composed of a great number of concentric spherical 
shells. By Cor. 1, the shell external to A x will produce no 
effect and the whole attraction will be due to the inner sphere. 
Now the law of attraction is 

F -f • Ei.5... i* R3 "j*i* 

r 81 • r u - ~ • — • • — — : • . -K. . r. 

S t 2 S 2 2 R2 f2 

273. — Cor. 4. At the surface of spheres having the same den- 
sity, gravity varies as the radius of the sphere. 

The law as before is 

F gl : F g2 :: — ■ :— - which becomes for Fig. 180, 

s, 2 s 2 

F . F .. m '. m * .- 3- R3 A*** ..-R. r 



gl • X So 



s, 2 s, 2 R 2 r 2 



274. — Deviating 1 Force. The force (F d ) perpendicular to 
the tangent (§265) and directed towards the centre of curva- 
ture produces deviation from the tangent, and may be called 
the deviating force. It is sometimes called centripetal force. 
See Newton's Second Law, §261. 

275. — Prop. The deviating force equals the product of the mass 

v 2 
of the body and the rate of deviation, or, F d = m — . 

9 

276. — Cor. 1. The preceding equation may be transformed 

v 2 o 
into F d : W :: — : - , that is, the deviating force is to the weight 
2g 2 

of the body as the height due to the velocity v, (§206) is to half the 
radius of curvature of the path. 



io8 

277. — Cor. 2. If a> is the angular velocity (§221) of a rotating 

W w 2 r 2 _ W 
body, F d = — — co 2 r, that is, the deviating force equals 

the continued product of the mass of the body, the square of its an- 
gular velocity and the radius of the path. 

278. — Cor. 3. If the body revolve uniformly in a circle and 
n equals the number of revolutions, then co = 27m and Fd = 

W 

~(47r 2 n 2 r) that is, the deviating force equals the mass multiplied 

9 

by 47r 2 n 2 r. 

279. — Cor. 4. If the time t of a single revolution is given 
Fd =^( 4 -)r,thatis,&c. 

280. — Newton's third law shows that when a body has its 
direction changed it offers to the change a reaction equal and 
opposite to the force tending to produce the deviation ; this 
reaction has sometimes been called centrifugal force. There is 
no more propriety in giving a separate name to this reaction 
than there would be to dignify by a special term the reaction 
which all bodies offer to being moved towards the centre of 
the earth by the action of gravity. 

Illustrations of the so-called centrifugal force are shown 
when a pail of water is revolved rapidly in a circle, the water 
pressing upon the bottom of the vessel and not falling out 
though inverted in part of its path ; by the communication of 
motion to grain by a horizontally rotating millstone, the grain 
being fed into the centre of the stone ; by the flying asunder 
of grindstones, emery wheels and millstones from too rapid 
rotation, the force becoming greater than the force of cohesion 
between the particles of the stones ; by a horse and an 
equestrian leaning towards the centre of the circular path in 
turning a corner ; by the elevation of the outer rail on a curve 
in a railroad track ; by the centrifugal clothes drying machine; 
the apparatus for separating crystallized sugar from molasses; 
by the potter's vessels on the revolving wheel ; by the gover- 
nor of the steam engine ; by the "centrifugal" railroad ; by 
the bulging form of the earth, and many common appliances. 



109 

It is obvious (§265) that if there were no deviating force the 
body would move in the direction of a tangent to the path by 
reason of its inertia. (§258). Various illustrations may be 
given of this tendency : the tangential path of a stone leaving 
a sling, of water flying from the periphery of a water wheel, 
of mud from a carriage wheel. 

The laws of deviating force may be experimentally verified 
by means of the "whirling table." 

281. — Prob. Find the resultant deviating force of any number 
of small bodies rotating around a fixed axis with angular velocity 
co, the bodies all being in one plane. 

W W W 3 

In Fig. 181 let m= — 3 ; m„ = - — -, m,= — , &c, be the 

9 ff 9 

masses of the bodies. Assume the rectangular axes Ox, Oy, 

Oz. Let the deviating forces be respectively F d] , Fa,, F da , &c. 

Let r n r 2 , r 3 , &c, be the distances of the centres of gravity of 

the small bodies from the axis Oz Let n a , d 3 be the angles 

the directions of the deviating forces make with the axis Ox. 

Then 

_ v x 2 W, v 2 W x <o% 2 W, 

F dl = m, - 1 - = — ■ — = — J ■ — L = — • co%. 
r, g r, g r 1 g 

v 2 2 W 2 v 2 2 W, fl» a r, 9 W 2 

F<j, = m 9 — = — ■ — = — ■ = — a> 2 r . 

2 r 2 £ r 2 g r 2 ^ 2 

Resolve F dl into components X, and Y,_ acting along Ox 
and Oy 

x W, x, W, 

X, = F dl cos. e x = F dl -j = ~ «>% "r 1 = 'J C ° 2 ' Xl 



y, W, y - W 

Y, ==F dl sin. & 1 = F dl . — = — co 2 v, ..Zi=-^, 1 

Resolve F d2 into X and Y 2 . 

Y 2 — P d> sin. 0, -= F d2 . — = — w 2 r 2 ■ - — — ^ 

r 2 # r 2 # 

Now, find the resultants of the X\s and Y's. 



■y, 



y. 



no 



W, 

9 



W W /W 

X =X 1 -fX > + &c.= — io'.x^— to* • x 2 + &c.=w 2 — l X] 

g g \g : 

x 2 +&c. X 



W, 



9 



w, w, 

Y^=Y 1 + Y 2 +&c=-^ 2 -y 1 + — <o 2 .v 2 +&c. 
9 9 

y> + &c/ 



w, 

~ y, + 



The resultant of X and Y is 

R d = i/XH-Y 2 -f &c~ 



:. = co- -\l 



W, 



x i 



g 



x 2 + &c. + 



' W > - W * MR 

— y l + ~r y 2 + &c. 



that is, the resultant deviating force 



equals the square of the angular velocity multiplied by the 
square root of the sums of the squares of the sums of the moments 
of the weights of the bodies relatively to two rectangular axes. 

Cor. Let x and y be the co-ordinates of the centre of gravity 
of the system of bodies. Then by the principle of moments, 



-/W, W 
+■ 



H 



a 

w, 



\g g 

Hence, R d 



+ &c.) = -y, 



w w 

— 'X,+— 'X. + &C. 

9 9 



W. _ \ w, 



(0 



(VP 



9 

% 
9 



9 



y. + &c. 



W, 
9 



1 1 



+ &c. i]°]=«.*V(* '+ y) (j+j +&c - 



w, w, 
'7 + 7 



v<^+ 



W,,. \' ,,w, , w, 

h &C. I = CO 2 I 



— + &c. r. But if the bodies were 
9 \ 9 9 I 

concentrated at their common centre of gravity this expres- 
sion would be the deviating force, that is, 

The resultant deviating force is the same as if the small 
bodies were concentrated at their centre of gravity. 



Ill 



_ y 



The direction of the resultant force is tan. R d Ox = -£; that 

x 

is, it is through the centre of gravity of the system. 

282. — Prob. Find the resultant deviating force due to several 
bodies rotating about an axis. 

w, w 2 w 3 

Let m,^— ; m 2 = — ; m 3 = — , &c. , be the masses of 

y y y 

the bodies (Fig. 182). Let co be the angular velocity of rota- 
tion, then the deviation forces of the bodies around the axis 
Ozwill be, if r n r 2 , r 3 , &c, are the distances of the centres of 
gravity of the bodies from Oz ; 

Wi w 2 w 3 

F d = - o>x ; F d = — a>*u ; F ds = — to*n ; &c. 

y y y 

Resolve each of these forces into components parallel to the 
axes Ox and Oy. 

Xi = F dl • 

X 2 = F dj • 
X 3 == F ds • 
Y, = F d , • 
Y 2 = F d2 • 



Y 8 = F 



Xi 


Wi 


Xl W, 


- — 


= 10 2 Ti 


• = CO*X x 


I", 


9 


r * <7 


_Xj? 


w * . 


x 2 W 2 




= — «> 2 r 9 


— W « X 


r 2 


9 


r 2 jf 


x s 


W 8 


X 3 Ws 


— 


= — fti a r. • 


— (0 2 Xo 


r 3 


9 


r 3 


y, 


w, 


yi w, 2 




— — w 2 r • 


— = — ■ w yi 


r, 


9 


r, ff 


y 2 


w « . 


y, w 2 2 




= — w 2 r • 


— = — «> 2 y a 


r 2 


9 2 


r 2 g 


y 3 


w * . 


y 3 W« 2 


: 


= — o> r 8 ' 


— = — io Vs 


r 8 


ff 


r 3 ■ 



Find, now, the resultants of the X's and Y's : 

W W 2 w s 

X=X I + X„+X,+&c.= — , «i 9 x i -f— oi-x, -{- — ^ 2 x 8 +&c. 

W Wo w 

Y=Yi +Y 2 +Y 3 -h&c.= — fli-y, + — ' w a y, + — ! Vy 8 +&c. 

The two preceding equations give the magnitudes of the 
resultants parallel to Ox and Oy. 



II 2 

The points of application of X, and Y, may be found by the 
principle of moments as follows: Let A n A 2 , A„ (Fig. 183) on 
the axis Oz, be the points of application of Xi, X 2 , X„ &c.,, 
Y„ Y 2 , Y s , &c Then if A x is the point of application of the 
resultant X and A y that of the resultant Y and O the centre 
of moments 

X • OA x = X, ■ OA, + X, • OA 3 -hX 3 -OA 3 -X 1 - Zl +X 2 • z 2 + 
X 3 • z 3 + &c. 

~ A m 1 x 1 z 1 + m 2 x 2 z 2 -hn,x : ,z :t + &c. 



Vilx 


IH] 


x^tr^x. 


,-fm, 


1 X 8 


-h&c. 


Si mi 


larly, 










OA y 


m iYi z 


:+in 2 y 2 z 


2 + m 


•y. 


z 3 + &c. 



From these equations it appears that the resultant is usually 
not a single force, for the points of application of X and Y are not 
the same. 

283. — Cor. If the centre of gravity of the rotating body or 
system be at the origin of coordinates, by reasoning similar to 
that in the preceding proposition it may be shown that the 
resultant deviating force about an axis is the same in magni- 
tude aiid direction as if all the bodies were concentrated at 
that point. The point of application will not necessarily be at 
the centre of gravity. 

284. — Cor. //"the axis of rotation pass through the centre of 
gravity of the body or system the resultant force equals zero y 
and there is no tendency to pull the axis out of place. 

285. — Cor. It also follows that unless X and Y are each 
equal to zero that a couple will result which will tend to change 
the angular position of the axis, notwithstanding that it 
passes through the centre of gravity. This couple may be 
found as follows : 

Let X and Y (Fig. 184) with the common point of applica- 
tion A be the resultant forces as already found. Since each 
acts at a distance OA from O by §64 



"3 





Force at O. 


Couple Around 


Couple Around 






Ox. 


Oy. 


X o 


X 




XOA 


Y o 


Y 


YOA 





Construct a new figure (Fig. 184a) in which OM y represents 
the moment, .position of the axis and direction of the first 
couple and Om x the second couple in the same respects. 
Complete the rectangle and the diagonal OM will represent 
the resultant couple in magnitude, direction and position of 
the axis. 



M = 1/ M= + M^ 

M 
Also, tan. f = ~ 

286. — It follows from these facts that in order that Oz may 
be perfectly at rest, that is, a " permanent axis," it is not suffi- 
cient that it should pass through the centre of gravity — that 
is, that the resultant centrifugal force equal zero, but it is 
also necessary that the sum of the moments in reference to the 
two rectangular axes should equal zero. 

Centrifugal Force Due to the Earth's Rotation on its 
Axis. 

287. — Prop. I. The centrifugal force at any point on the eartJi 1 s 
surface equals the centrifugal force at the equator multiplied 
by the cosine of the latitude of the place. 

Let A (Fig. 185) be any point on the earth's surface, EQ 
the equator, NS the axis. Let the earth rotate with" angular 
velocity to. Call the linear velocities at A and the equator v a 
and v e , and the distances of E and X from the axis respec- 
tively r e and r a . Call the angle ACE or the latitude of the 
place X. Then the centrifugal force at A on unit mass placed 
there is 

v 2 • r ' 2 

_ V a 2 — —7^" _ Ve * ' r * 2 _ V * T * _ Ve" ' T e COS. / _V^ 

a 7~ r e ~ r 2 • r r 2 ~ r 2 _ r 

r a 1 e * a Ie e e 

r a 

c . v e r a 

Since v e : v a :: ^r e : wr a :: r e : r a or v a = ~ 

r e 



"4 

Also r a = r cos. /. But — is the centrifugal force at the 

r e 

equator, hence the truth of the proposition. 

288. — Prop. IT. The component of the centrifugal force at any 
point on the eartli s surface opposed to the attraction of gravi- 
iation at that poiitt equals the centrifugal force- at the equator 
multiplied by the square of the cosine of the latitude. 

Resolve F (Fig. 186) into two components AN in the di- 
rection of the radius r e and AT tangent to the surface at A. 
Then AN = F da cos. / = F de cos. 2 I 

At the equator where X is zero this component is the great- 
est possible and it diminishes toward the poles at which places 
it becomes zero. 

289. — Prop. III. The tangential component of the ce?itrifugal 
force at any point is equal to the centrifugal force at the equa- 
tor multiplied by half the sine of twice the latitude of the place. 

* ~. _ sin. 2 / 
AT == F da sin. / = F de cos. I sin. / = F de 

If the particle is at the equator I = and AT = 0; it then 
increases until at 45 it is a maximum ; it then diminishes 
until the pule is reached where it is zero, since X = 90 . 

This component tends to move the particles towards the 
equator, and if the earth were plastic there would be an accu- 
mulation of matter at the equator until this component of its 
centrifugal force would be balanced by the component of 
gravity on the inclined plane at that point. 

290. — The apparent weight of a body is diminished at the 
equator on account of the greater centrifugal force there and 
also on account of the piling up of the matter due to the tan- 
gential component which makes the equatorial diameter 
greater than the polar. §269. 



r a 



291. — Prob. 1. The centrifugal force at the equator Fd e = 
- — o. 1 1 126 feet. 



"5 

Prob. 2. The centrifugal force at any other place may be 
found by multiplying the cosine of the latitude by .o. 1112. 

Prob. 3. The apparent weight of a body at the equator 
would be zero if the earth were to revolve 17 times as fast as 
it does now. 

Prob. 4. The apparent weight of a body at the equator is 
diminished 1-289 of its weight by the centrifugal force. 
g = 32.09022 at the equator. 
ff, =32.09022 -f 0.1 1 126 = 32.20148. 
W g .11126 1 

Wi~ g~~ 32. 23 148 "289 

292. — Prob. 5. How high must the outer rail of a track be 
raised to balance the centrifugal force due to the velocity v 
of a train of cars, the radius of the curve being r and the dis- 
tance between the rails s. 

Let AB (Fig. 187) be the plane of the horizon; AC the 
slope ; BC=h the height of outside rail necessary to balance 
the centrifugal force of the train; d=the distance between the 
rails and the angle of inclination of the slope to the horizon. 
It is evident that the resultant of the weight W of the train 
and the centrifugal force F d must be perpendicular to AC. 
Then F d : W :: gF d : gW :: WN: gW :: l: tan e*. 

Wv 2 

— :W::l:tan. V. 

9 

V 2 

. • . tan. = — 
9* 
but h = d tan. (nearly) 

.'. h = dX 

9* 
293.— Simple Harmonic Motion. If a body m, revolve uni- 
formly in a circle of radius r (Fig. 188), with angular velocity 
co the deviating force may be represented by F dl . If F dl be 
resolved into two components X x and Y ] along two rectang- 
ular axes with O at the centre of the circle as origin, Y 1 =m 1 
a>2y n that is, the body is urged towards O by a force which 
varies as the distance from O. 



u6 

If another body m/=m, be imagined to start from a at the 
same time with m n m/ will vibrate along ac, always keeping 
on a horizontal line x with m i5 since m/ is acted upon by a 
force, irij^Vj directed towards the centre O. 

The distance of the line x from Ox is y = r cos. cot — r cos. 
if t is the time measured from the instant when the moving 
body is at a and B = tot. 

The velocity of m/ is the same as the component velocity 

dy 
of m 1 parallel to Oy, that is -— = — cor sin. cot == — cor sin. 0. 

Since mi describes the whole circumference of the circle 
while m/ moves to and fro along ac, the time of a double 
vibration is the same as the time m, takes to revolve in the 
circle or if t equals the time of a single revolution and n the 
number of revolutions per second 

t — - — — — 2 a \f — §277. Fd T is the greatest value 
n co V ^F dl 

the deviating force corresponding to r, can have, m/ is said 
to have a simple harmonic motion along the axis yy. 

294. — The Pendulum. Definitions. A simple pendulum is 
a material particle attached by an inextensible and weightless 
rod to a point of suspension about which it is free to move. 
The path is therefore circular. 

A single vibration is the motion from the highest point on 
one side to the highest point on the other, or its equivalent. 

A double vibration is a to and fro motion. 

The amplitude is the extent of the swing measured by the 
angle between the extreme positions. 

The length of a simple pendulum is the distance from the 
point of suspension to the particle. 

The time of a vibration — single or double — is the time it 
takes to describe either. 

295. — Prop. The time of a double vibration of a pendulum is 
equal to the quotient obtained by dividing the square root of the 
length of the pendulum by the square root of the acceleration due to 
gravity, multiplied by 2n. 



ii 7 

Let OB (Fig. 189) represent a simple pendulum. The only 
force acting on it to produce motion is the weight g W. 
Resolve this force into two components, gT tangent to the direc- 
tion of motion and gN at right angles to the tangential compo- 
nent. ThengT=gW sin.gWT=gW cos.TgW=gW sin.goz= 

o-p 
gW • - — . If the arc gP, be made very small, the chord gP 
?■> 

s 
becomes equal to it and gT = gW- -, in which 1 equals the 

length of the pendulum and s the distance of the particle from 

P,. In this equation gW and 1 are constant, hence gT or the 

force which produces the motion varies as the distance from 

the fixed point Pj and the motion of B is therefore the same 

s gW 1 
as that of m/ in §293 and the torce equals F d) =gW- -or %—=- 

1 ra, s 

Hence the equation of time of §293 will apply to this case, 
namely : t = 2n -x ; in which F d is the maximum cen- 

trifugal force corresponding to the half amplitude r. Substi- 

. . W 1. t . 

tuting the ratio -=; — = -111 this expression 

•F dj S 

f = 27T -\ — — 2~ \l — = time of double vibration. 

\<7 S v g 

296. — If any two of the three quantities, t, 1 and g are given 
the other may be found. 

Prob. Find the length of a pendulum which will beat sec- 
onds in latitude 45 where g = 32. 171; also at London where 
g = 32.182 ; at the equator where g = 32.088 ; at the poles 
where g = 32.253. 

Prob. Find the value of g or the acceleration due to grav- 
ity if t and 1 are given, t may be found by counting the 
number of vibrations in a given time and 1 by actual meas- 
urement. 

The value of g at any place may be determined approxi- 
mately by the following formula: 



n8 

g = 32.088 [1 -(-. 005133 sin. 2 /] (1 — — J in which X is the 

latitude of the place, h its height above the sea level and R 
the radius of the earth or 20.900000 feet. 

297. — This equation (§295) shows that the time of vibration 
of the pendulum depends only upon the length of the pendu- 
lum and g. The material, length, form and weight exert no 
influence. If the arc is very small the time is independent of 
the amplitude, a fact discovered by Galileo and called the 
isochronism of the pendulum. 

298. — Comparison of Times of Vibrations of Pendulums 
of Different Lengths. 

If Tj is the time of a double vibration of a pendulum whose 

length is L, then, T, = 2* ^—^ ■ also if T, is the time of a 

double vibration of another pendulum of length L 2 then, T.. 
= 27T -- 2 from which it follows that 

T • T :: — — : -7=1 that is, the times of vibration are as the 

1 ' g> v g* 

square root of the length divided by the square root of the acceleration 
due to gravity. 

Cor. 1. If the pendulums are caused to vibrate at the same 
place, that is, where g 1 = g^ then, 

The times of vibration are as the square roots of the lengths. 

That the time of vibration varies as the square root of the 
height may be experimentally verified by taking three pen- 
dulums of length 1, 4 and 9 feet. If the numbers of vibrations 
are counted in a given time they will be as 3, 2, 1. That is, 
the number of vibrations varies inversely as the square root of 

11 11 7 /— 

the length or N, : n 9 :: -7= : /p or n" : n" :: V ^ : **> but 

time of a single vibration evidently varies inversely as the 

number of vibrations or T, :T 2 : — - : ^hence T s : T 2 :: i/L 1 '.y L a 



ii 9 

Cor. 2. If the pendulums are of the same length the times of 
vibration are inversely as the square root of the acceleration due to 
gravity, zvhen made to vibrate at different places. 

299. — Prob. Given the length of a pendulum which beats 
seconds, find the length of a pendulum which will beat half 
seconds ; also one which will beat once in two seconds. 

300. — Definitions. The Compound Pendulum is the one 

in ordinary use. It is a massive body of various forms. 

The axis of suspension is the horizontal line about which it 
vibrates. It is vertically above the centre of gravity of the 
pendulum. 

The axis of the pendulum is a line drawn through the centre 
of gravity at right angles to the axis of suspension. 

Since the compound pendulum is made up of a great num- 
ber of simple pendulums rigidly connected, it is obvious 
(Fig. 190) that those particles at equal distances from the axis 
of suspension will move with equal velocities ; those nearer 
the axis with velocities less than the whole body and those 
farthest away with velocities greater. Also the momenta 
of the upper particles must be less than the lower. There 
must be a point where these momenta balance each 
other. This point is called the centre of oscillation. The cen- 
tre of oscillation cannot coincide with the centre of gravity. 
In the ordinary form of the pendulum it must be below the 
centre of gravity. 

The axis of oscillation is a line drawn through the centre of 
oscillation parallel to the axis of suspension. 

The length of a compound pendulum or the equivalent simple 
pendulum is the distance from the axis of suspension to the 
axis of oscillation. Varying the position of the bob of the 
pendulum changes its length by changing the position of the 
centre of oscillation. 

301. — [nterchangeability of the Axis of Oscillation and the Axis 
of Suspension. 

Huygen discovered that if the pendulum is first suspended 



120 

from the axis of suspension and the vibrations counted in a 
given time and then inverted and suspended from the axis of 
oscillation the time of vibration is unchanged. 

This property furnishes an accurate method of determining 
the length of the equivalent simple pendulum corresponding to 
any compound pendulum ; for, if two moveable knife edges 
are attached to the pendulum, and the pendulum first sus- 
pended from one and the number of vibrations in a given time 
counted ; then suspended from the other — which is moved 
until the number of vibrations in the same time is exactly the 
same- — the distance between the knife edges can be measured. 
This distance is the length of the compound pendulum. Such 
a pendulum is called Kater's. 

302. — Applications of the Pendulum. 1. To measure time. 
Before the introduction of the pendulum time was measured 
by the uncertain methods of allowing water to drop from a 
vessel as in the clepsydra, or sand to pass from one vessel to 
another as in the hour glass, by the burning of caudles, the 
pulsations of the wrist, the beating of the hand as in music. 

The pendulum being isoschonous became in the hands of 
Galileo a valuable means of measuring time. Huygens of 
Holland first connected it with clock-work and thus kept up 
the vibrations by applying a force just sufficient to overcome 
friction at the axis of suspension and the resistance of the air. 
A pendulum which will beat seconds is of such length that it 

will indicate at each vibration of a meait solar day. 

86.400 

By shortening the pendulum it may be made to indicate the 

corresponding part of a sidereal day. 

303. — 2. To Determine the Form of the Earth. If a pendu- 
lum of constant length be carried to different parts of the 
earth's surface and the times of vibration observed, by §298, 

Cor. 2. T\ : T„ :: -7=- ■ — or T, 2 : T 2 2 :: — : — ; but (§269) 
" V g x Q, ft ft 

F g] :F g2 ::^:^:: — : — therefore T, : T 2 :: s t : s 2 that is the 



121 

times of vibration are as the distances from the centre of the 
earth. 

Kater, Sabine and others showed by such measurements that 
the earth is an oblate spheroid. 

Measurements with the pendulum show that the equatorial 
diameter is thirty-four miles greater than the polar, while ac- 
tual measurements give only twenty-six miles difference. 

It is evident that with this proposition and experiments 
with the pendulum the law of gravity (§269) mighthavebeen 
demonstrated. 

304. — As a Standard of Linear Measure. 

The yard (§173), theoretically, could be replaced if the ratio 
of its length to that of a pendulum beating seconds were 
given. 

Now, the ratio of the length of the yard to the length of the 
pendulum beating 86, 400 times in one mean solar day 'in vacuo, 
at the sea level, in the latitude of Greenwich, is 39.13929 
inches of Bird's standard of 1760 at 62°F. This ratio was 
fixed by Capt. Kater. 

305. — The Conical Pendulum is a ball suspended from a 
fixed point by means of a rod, the weight of which may be 
neglected when compared with the weight of the ball. The 
ball revolves in a horizontal circle and the rod marks out a 
cone. 

B (Fig. 191) represents the ball suspended from the point 
O. It is evident that, when in motion, the weight W of the 
ball, the centrifugal force F d and the tension T on the rod 
must balance when the path described has a constant radius. 
Let r be the radius of the circular path, then 

Fd : W :: sin. OgW : sin. OgF d : sin. (180 — S) =sin. 6 : sin. 
(90—0) = cos. S. 

F d =W • - 1 — = W tan. 6 ; but 
cos. 6 

T*r tt7 4~ 2 r 2 TTT 47T 2 h 2 tan. 2 # „ r - . a 
v _ W v 2 _ W - W - — : _ W 47T 2 h tan. d 

d ~ir "7;._!|_=7-_^i_~7- -?— 



122 

Solving for t, t= 2' a/—. §295 

y 
Hence, the time of revolution of a conical pendulum equals the 
time of a double vibration of a simple pendulum whose length is h. 

306. — The Theory of Atwood's Machine, Let A (Fig. 192) 
be a smooth peg or frictionless pulley over which a cord is 
run, having weights W, and W 3 attached to the two ends. 
Let W, be greater than W 2 . It is obvious that W, will move 
down and W 2 will be drawn up. According to the second law 
of motion F = ma ; in this case the moving force is the dif- 
ference between W, and W 2 or W, — W 2 ; the mass moved is 

w > + w. i r w W w > + w 2 w-w a 

; therefore W, — W,= " ~ a ora=-^^, 

that is, the motion is like that of a free falling body but the 

W- W 2 
acceleration is diminished in the ratio of 1 : ^rr^rr- Hence, 

Wj+Wa 

all the equations of (§204) apply to this case if ^-77™/ 9, be 
substituted for g. 



123 



THE SCIENCE OF ENERGY. 



307. — Work. Definitions. Work is the production of mo- 
tion against resistance. 

Resistance is a force which acts opposite to the direction of 
motion tending to check the motion. 

Work is measured by the product of the resistance and the dis- 
tance through which the point of application of the resistance is 
moved. If F r is put for the resistance and s for the distance, 
Work = F r s = mas. §263. 

Suppose the point of application A (Fig. 193) of the force 
F r to be moved from A, to A a . Resolve AjA 2 into two compo- 
nents A 2 A 3 and A 1 A 3 = A X A 2 cos. 6. Then F r ■ A X A 3 or F r . A x A a 
cos. 6 is the work performed in moving A against the force F r 
from A j to A 2 . 

It is obvious that the same result could be obtained by re- 
solving F (Fig. 194) into components along and at right angles 
to A,A a . The work would then be F r • A, A, = F cos. 6 • AjA. 

Hence, if the force is oblique to the direction of motion 

308. — Work equals the force multiplied by the component of the 
displacement along the line of action of the force or the displace- 
ment multiplied by the component of the force along the direction of 
the displacement. 

309. — Work is performed when bodies are lifted above the 
surface of the earth ; for example, a pumping engine performs 
work, in lifting water to a reservoir, equal to the weight 
of the water multiplied by the distance through which its cen- 
tre of gravity is lifted ; work is performed in winding up the 
weight of a clock ; a locomotive performs work against grav- 
ity in ascending a grade ; a stationary engine in lifting the 
weight of a pile driver. 



124 

In the case of work performed against gravity F r becomes 
the weight of the mass moved, or F r s = Ws — mgs. §266. 

310. — When a body is deformed, as in bending a bow or 
stretching a piece of rubber work is performed. Heat performs 
work in expanding bodies ; electricity performs work in de- 
composing water against the force of chemical affinity, and 
the molecules of hydrogen and oxygen are in the same condi- 
tion as a body, and the earth when separated from each other 
after work is performed in lifting the body. 

311. — The following consequences proceed immediately 
from the definition of work. 

1. A free body moving uniformly performs no work. §258. 

2. A man who merely supports a load performs no work. 

3. If the force is perpendicular to the direction of motion 
no work is performed against it. Hence, a body moving on 
a level surface perforins no work against gravity. 

4. The actual path described by a moving body is a matter 
of indifference. The same amount of work will be performed 
against gravity in taking a zig-zag path up a hill as in going 
vertically upward. The work performed in raising a heavy 
body along a smooth inclined plane is the same as would be 
necessary to raise the body through the vertical height of the 
plane. In Fig. 195, if 1 is the length of the plane, h its height 
and d its angle, F r s = F r l = W sin. d 1. But this expression 
equals Wh. 

312. — Unit of Work is obtained when F r s == mgs = l ; that 
is, when F r = 1 and s == 1, or when unit force acts through 
unit distance. 

In the C. G. S. system (§264) unit force is the dyne. A dyne 
acting through one centimetre is the unit of work called the 
Erg y or since F r s — mgs == 1, the unit of work will be per- 
formed when — — - gramme is raised through one centimetre, 
981 

since . q8i • 1=1. 

981 * 

The megalerg— 1,000,000 ergs. 



125 

The ergten = 10 10 ergs = 10,000,000,000 ergs. 

If the pouudal (§264) is taken as the unit of force 32.2 foot- 
poundals will be required to lift one potmd-mass through one 
foot. F r s = nigs = 1 X32.2 X 1 = 32.2 foot-poundals. 

If a pound is taken as the unit of resistance and a foot as the 
unit of distance a foot-pound will represent the unit of work. 

A foot-pound 'is the work performed in lifting a pound ver- 
tically through one foot. 

If 50 pounds are lifted through 10 feet 500 foot-pounds of 
work are performed. The work would be the same if 500 
pounds were lifted through one foot or one pound through 
500 feet. 

Since g varies from place to place the foot-pound is a varia- 
ble quantity. Notwithstanding this objection the foot-pound 
is usually employed by British and American engineers. It 
may be reduced to absolute units by using the equation F r s= 
Ws=mgs ; hence, the foot-pound equals g foot poundals. 
The foot-pound equals 13,562,691 ergs. 

The kilogramme-metre is the French unit of work and rep- 
resents a kilogramme raised through one metre. 1000 grms. 
X 100 cms. X g = 98, 100,000 ergs. 

Rate of Work or Activity is the amount of work performed 
in unit time. 

An arbitrary unit of activity in common use is the horse- 
pozver, which represents the work performed in lifting 33,000 
pounds through one foot in one minute, or 550 foot-pounds 
per second. 

The French horse-power (cheval-vapeur) is 75 kilogram- 
metres or 7,500,000 g — 7,357,500,000 ergs per second. 

313. — Work may be represented by an area. 

If the resistance is constant the work performed may be rep- 
resented by the area of a rectangle, one side of which repre- 
sents the distance over which the resistance is overcome and 
the adjacent side the given resistance. 



126 

If the resistance is variable the work may still be repre- 
sented by an area. On a horizontal line (Fig. 196) take suc- 
cessive lengths to represent distances through which the re- 
sistances act. Draw at the extremities of these lines vertical 
ordinates representing the resistances. Complete the rectan- 
gles. The area of each rectangle represents the work done 
by the corresponding force, The total work will be repre- 
sented by the sum of all the rectangles. 

If the resistance be imagined to change very gradually a 
diagram like Fig. 197 will result and the sum of the elemen- 
tal areas will be the whole work done. 

The area can be approximately calculated by Simpson's 
rule 

Take the sum of the first and last ordinate, twice the stint of 
the other odd ordinates, four times the sum of all the even ordi- 
nates, multiply the result by one-third of the common distance 
between two adjacent ordinates. 

Another useful rule is the following : Divide the figure into 
n vertical strips. Then i-nth of the mean of the first and last 
ordinates, plus the itttermediate ordinates, will give the mean 
ordinate. This ordinate multiplied by the length of the 
diagram will give the included area. 

The indicator diagram of the steam engine illustrates a case 
of variable pressure. 

314. — Prop. The work perfomed in lifting a system of bodies 
vertically through a given distaitce is equal to the weight of 
the system multiplied by the distance the common centre of 
gravity is raised. 

Suppose bodies whose weights are W„ W 2 , W 3 have their 
centres of gravity at the distances z„ z„ z s above a horizontal 
reference plane. The distance of the common centre of grav- 
ity from the same plane will be (§116) 

z= W»z, -f- W g z 2 -f „W 3 z 3 
w, + W 2 + W 3 

Let the bodies be lifted the additional distances z/, z/, z/ ; 



127 

then the distance of the common centre of gravity will be- 
come 

_, = W, (Z/ + Z x ) +W 2 (Z /+Z 2 ) -rW 3 (z,' + Z.) 

w, + w 2 + w 3 ~ 

Subtracting the former equation from the latter 

_, _ _ = W,z/+W 8 z/+ Wa' or 

W,+W,+W, 
(5' _ z) (W,+W,+W s ) = W,z,' + W A ' + W 3 z s ' 

The first member is the work performed in lifting a body 
equal to the sum of the bodies concentrated at the centre of 
gravity ; the second the sum of the separate works performed 
in lifting each body through the distance over which its cen- 
tre of gravity is moved. 

315. — Prop. When a pressure acts upon a piston in a cylin- 
der the work performed is equal to the product of the inten- 
sity of pressure, (§14) the area of the piston, and the length 
of the stroke or the intensity of pressure multiplied by the 
volume traversed by the piston. 

Let A be the area of the piston, p the intensity of pressure 
or pressure upon each square inch, and s the distance trav- 
elled by the piston or the length of the stroke. It is then ob- 
vious that F r s = (pA) s= p (As)= pV in which V is the volume 
traversed by the piston. 

316. — Prop. When a body rotates on an axis against a re- 
sistance applied at the circumference the work performed equals 
the moment of the resistance multiplied by 2x times the number 
of revolutions. 

For, calling F r the resistance as before, the distance traversed 
by a point at a distance 1 from the axis in one revolution 
will be 2~1 ; for n revolutions 27rln. Hence, F r s = F r (27ml) 
=F r l (27m) or the moment of the resistance multiplied by ztl 
times the number of turns. 

3 1 7. — Prop. The whole work perfor7ned at various points in 
a system of moving bodies equals the sum of the works at the 
separate points. 



128 

Let F ri , F r2 , F rs be resistances applied at various points of a 
moving system and s n s 2 , s 3 , the distances through which the 
points of application of the resistances are moved, the whole 
work is obviously equal to F ri Sj -f F r2 s 2 -f F r3 s 3 . 

Cor. If the body or bodies simply have a motion of transla- 
tion (§231) then s, = s 2 = s 3 — s and the whole work equals 
(F ri + F r2 + F r3 ) s. 

Cor. If the pieces rotate then the whole work will equal 
Fr-1, 2-n, -f- F r2 l 2 2~n 2 + F r J 8 2-n 8 . 

Cor. If there is a single rotating part 27m, = 2;rn 2 = 2~n 3 = 
22m and the work becomes 2~n (F^^ -f- F r2 l 2 + F rs l 3 ). 

318. — In any machine each resistance at any point may be 
reduced to an " equivalent resistance at the driving point." 
It is always possible to determine the ratio of the velocities of 
different pieces of a machine from their mode of connection. 
For example let v d be the velocity of the driving point and 
Vrj the velocity of a resistance at any other point, and suppose 
Vd : v ri :: 1 : n determined from the mode of connection of the 
parts. Multiplying the first two terms by F ri the proportion 
becomes F ri ■ v d : F ri • v ri :: 1 : n or F ri • v^ = nF ri • v<j, which 
equation shows that any resistance F r , moving with a velocity v r , 
equals n times the same resistance at the driving point moving with 
the velocity of the driving point. 

Cor. The equivalent resistance at the driving point for all 
pieces will be the sum of the individual equivalent resistan- 
ces ; e. g., rijFr, + n s Fr a + n 8 F r , -h &c, and the work per- 
formed will equal the equivalent resistance at the driving 
point multiplied by the displacement of the driving point. 

319. — 111 a train of mechanism consisting of rotating pieces 
the "equivalent moment of resistance" at the driving axle 
"may be obtained. For example, let F r , at the extremity of an 
arm \ represent a resistance ; also, let co ri be the angular ve- 
locity of the resistance and co d be the angular velocity of the 
driving axle. From the mode of connection the ratio between 
these two angular velocities may be found; e. g. (o Ti : co d :: n : 1. 
Multiplying the third and fourth terms by F ri l, the proportion 



129 

becomes to Ti ;w d :: nF r , 1, : F^, or o> ri (F ri l x ) = co d (nF,,!,), that 
is, nF ri lj moving with the angular velocity o>d at the driving 
axle is equal to F r , 1, at the rotating piece moving with its ve- 
locity. After having found the "equivalent moment of re- 
sistance' ' for each piece, the whole moment of resistance may be 
determined by taking their sum and the whole work by mul- 
tiplying by the angular displacement of the driving axle. 

320. — Work of Acceleration of Translation and Rota- 
tion. 

When the velocity of a body is accelerated work must be 
performed to produce the acceleration. 

321. — Prop. The work necessary to increase the linear velocity 
of a body depends only upon the initial and final velocity and is 
equal to the increase of kinetic energy due to the final velocity. 

Let a body of weight W have a uniform acceleration a. The 
second law of motion (§261) shows that a force varies as the 
change of momentum it will produce. Let F r be the resist- 
ance which the body offers to having its velocity accelerated, 

W 
then F r : W :: a : g and F r = — a. The third law (§267) ena- 
bles us to substitute the resistance to motion, for the force 
producing the motion. As before, F r s is the expression for 

work, then F r s = (™a )s^ (*=*) (*±* t]^ 
\g I g \ t I \ 2 I 2 g 

. Wv 2 Wv 2 

(v.— vj a = — -' — x -, since (§192) v 2 =v, + at.'. a== 

v 2 — v, , , e ss Vo+v lv ~, . Wv 2 V 2 

— and (§196 s -= -~ -t. The expression = m — 

t V ^ ' 2 20 2 

is called kinetic nergye. §332. 

If the acceleration be not uniform but variable the Calculus 
shows that the result is the same. 

Cor. When several works of acceleration are performed the 
total work is equal to their sum. 

322. — Prop. The work necessary to increase the angular 
velocity of a rotating body depends only itpon the initial and 
final angular velocities and is equal to the sum of the works 



130 

necessary to produce the angular acceleration of all the points 
in the body, or it is equal to the moment of inertia of the body 
multiplied by the increase of the square of the angular velocity 
divided by twice the acceleration due to gravity. 

Let B (Fig. 198) be a body rotating- around the axis O with 
initial angular velocity co t . Let its angular velocity be in- 
creased to o> 2 . Imagine the body to be divided into a great 
number of small parts. Then the linear velocity of one of 
them, W x (§222) will be v x = co 1 r 1 and its final linear velocity 
will be v 2 = ^ 2 r i ; also v 3 2 = at'r* and v 2 2 = fl^'r, 2 , which sub- 
stituted in the general equation for work of accelerated trans- 
lation (§321) gives — '-co 2 r 2 -■ a>*r* = — r 2 \°^ — I . For 

v ° /fe 2f/ ' l zg 2 * g \ 2 / 

another small part, W 2 , a similar expression may be obtained, 
namely, — 2 r 2 2 — - ) , &c. The total work will be the 

y \ 2 

sum of all the elemental works, or 

W, /ft> 2 — ft) 2 \ , W„ [co 2 —co 2 \ ,W /w, 2 -w, 2 l 



/ft) 2 ft) 2 \ TTT /ft) 2 CO ~\ , XTT /ft) 2 ft) 2 \ 



+&c. = 



tw 



f W^ 2 4- W.r. 2 + W r 3 2 -h &c. ) U 1A- = I • 

The first factor is the .«//# #/~ ^<? elemental weights each 
multiplied by the square of its distance from the axis and is 
called the moment of inertia of the body. It is usually repre- 
sented by I, 

323. — Prob. Find the work of acceleration of translation of a 
machiyie of several pieces reduced to the driving point. 

If Wi is the weight of a moving piece of a machine, v Wl its 
velocity, v d the velocity of the driving point, and the ratio of 
the velocities determined by the mode of connection v Wl : Vd 

:: n i: i, then ^^:^::n 2 : 1 ,or^ V i-^n 2 W 1 ^ that 

zq zg zg zg 

is,n 2 W is the equivalent weight at the driving point necessary 



I3 1 

to produce the given acceleration. Other pieces of the ma- 
chine will give similar expressions and all the weights re- 
duced to the driving point will be n x 2 W x + n 2 2 W 2 + n 3 2 W s 
or 2 • n 2 W. The whole work of acceleration is then 

V 2 V 2 

324. — Prob. Find the work of acceleration of rotation of a ma- 
chine of several pieces. 

The mass of each piece may be regarded as concentrated at 
the extremity of the radius of gyration. Then let v d be the 
velocity of the driving point at any instant, co 1 the angular 
velocity of the rotating piece. Then v 1 =co u o i and if v d : a> 1 p l 

co 2 2 - _. , 
:: n : 1. v 2 : co *p* :: n 2 : 1. n 2 =-L±J- m Substituting m the 
d v 2 

Y 2 tt 2 V 2 — V 2 CO 2 P 2 

expression — -2 n„W it becomes — I — t -^- W. 

325. — The work of a couple equals the force multiplied by 
the sum of the dista7ices the points of application of the forces 
are moved through in a given time, that is, Fs;but this sum 
is the arm multiplied by the angle through which it rotates 
about the axis in the same time. 

F(ld6>) = (Fl) d0 ■= Md0 = Mfl«it. 

326. — Energy.. Definition. Energy is the power to per- 
form work. There are two kinds — Static and Kinetic. 

327.— Static Energy, Potential Energy, Energy of Posi- 
tion, Energy of Separation, Energy of Stress. 

Work is performed upon a body to lift it above the earth's 
surface represented by m^s. If the body, after being lifted 
to a height is kept there, it possesses the capability of per- 
forming work by its fall, when released. If it be connected to 
a train of mechanism it will perform work in falling equal to 
the amount which was performed on it to place it in its eleva- 
ted position. A wound up clock-weight has the capacity to 
perform work by reason of its position. It has, therefore, 



132 

energy of position, possible energy or potential energy. Since there 
can be no motion without the exertion of the force of gravity, 
which acts between the earth and the body as a stress, the en- 
ergy is sometimes called energy of stress. Since the body is at 
rest when it has this potential energy, it is often spoken of as 
static energy or energy of a, body at rest. It is also called energy 
of separation, because it becomes possible only upon the separa- 
tion of the bodies produced by the performance of work. 

If a body has its configuration changed by the performance 
of work upon it, as in bending a spring, stretching a piece of 
india-rubber or compressing a gas, these bodies are in a con- 
dition of potential energy, and when allowed to recover will 
exert energy to an amount equal to the work performed to 
produce the change. The air gun, the clock-spring, a bent 
bow, represent mechanical arrangements for the storing up 
of energy. 

The same principles apply to the separation of masses, of 
molecules and of atoms. When a piece of iron is elongated, 
within the limits of elasticity, by performing work against 
the force of cohesion, the molecules have potential energy just 
as a body has when separated from the earth. The constitu- 
ents of gun-powder have potential energy by reason of the 
work performed in separating their atoms against the force 
of chemism. 

A conservative system is one in which if a change is pro- 
duced in its configuration by the performance of work, it will 
tend to resume its original configuration when released and 
restore the work performed upon it. 

An attraction is called negative ; a repulsion positive. 
If two bodies attract each other, the potential energy 
is nothing when they are separated to an infinite distance ; it 
will be greatest when the bodies are in contact. 

When the earth and another body are considered, the po- 
tential energy is for convenience said to be nothing when 
they are in contact. 

The work done by an attracting force to separate bodies 
against their mutual attraction must be a negative quantity. 



133 



If the force between two bodies is a repelling one the po- 
tential energy is a maximum when they are in contact and 
zero when separated from each other an infinite distance. 

Potential energy is expressed by multiplying the force by 
the displacement its point of application is capable of moving 
through in the direction of the force, Fs. The unit of poten- 
tial energy is the foot pound. §312. 

If the point of application is actually moved through a dis- 
tance, energy is exerted to an amount represented by the 
force multiplied by the distance moved. 

That the energy of a force is simply the force multiplied 
by the displacement in the direction of the force, may be 
shown by resolving the force and the distance along three 
rectangular axes, finding the energies along the axes and add- 
ing them. 

Suppose a point moves over a small distance s the direction 
of which makes the angles a 1} /9 X , p H with the three rectang- 
ular axes Ox, Oy, Oz. Let F be the force making the angles 
a -2) A) T*i with the same axes. Then 



Components of s 
Components of F 
Energy 

The sum of the energies = Fs cos. a 1 cos. a 2 + Fs cos. j3 i 
cos. /? 2 + F"s cos. y 1 cos. y 2 — Fs (cos. a x cos. « 2 + cos. ft cos. ft 
-b cos. y l cos. )\) = Fs cos. 0. Hence, 

328. — Prop. 'Energy may be regarded, if the force is 
oblique to the directio7i of possible motion, as the force multiplied 
by the component of the displacement in the direction of the 
force, or as the component of the force in the direction of the 
displacement. F (s cos. &) = (F cos. 9) s. §308. 

329.— -The energy of a couple is represented by the same 
expression as the work of a couple, with contrary sign. 

330. — Prop. The energy of a varying force equals the sum of 
the energies due to the different distances over which parts of the 



Along Ox. 

s cos. « x 

F cos. « 2 

Fs cos. «j cos. « 2 


Along Oy. 

s cos. ft 

F cos. ft 

Fscos.ft cos. ft 


Along Oz. 

s cos. y 1 

F cos. y 2 

Fs cos. y x cos. y 2 



*34 

force regarded as uniform act. Let F, be the force acting over 
the distances Si, F 2 over s 2 , F 3 over s s , then the energy is 
F 1 s 1 + F 2 vf-F s s 3 . 

331. — It is evident that for every case under work there is 
a corresponding expression under potential energy, the only 
difference being in the sign. §307. 

Energy like work may be represented by an area. 

332. — Kinetic Energy, Actual Energy, Energy of a Body 
in Motion. 

Bodies in motion have the power of performing work. A 
cannon ball could be so arranged that when it was fired from 
a gun it would perform work in lifting another body. Run- 
ning water received on a wheel performs work through the in- 
termedium of mechanism. A bullet fired vertically upwards 
performs the work of raising its own mass against gravity. 
By §208, (4) if its velocity is v it will rise to a height repre- 

v 2 
sented by — , hence, the whole work performed is m^h = mff 

— -= . This expression is called the Kiiietic Energy, 

Energy of Motion, ox Actual Energy of the moving body. It 
depends simply on the mass moved and its velocity. 

W v 2 

Since the mass of a body equals — , kinetic energy m — 

v 2 
may be regarded as W — or the weight of a body multiplied 

2 ff 
by the height from which it must fall to acquire its velocity v- 

The unit of kinetic energy is a foot-pound since it is the 
product of a weight by a height (§312). mv 2 was formerly called 
vis viva, hence the kinetic energy of a body is one-half of its 

vis viva. 

333. — The kinetic e?tergy of a body is independent of its di- 
rection of motion, for if its velocity v be resolved into compo- 
nent v x , v y , v z , along three rectangular axes the kinetic en- 

V 2 v 2 v 2 

ergies due to these velocities will be m — , m — , m — , and 

222 



135 

their sum will be m — + m— ^--f in— =m x y z l— m 



2 2 2 \ 2 / 2 

the same expression as before. 

334. — Prop. The total kinetic energy of a system of bodies, each 

moving with its own velocity, equals the sum of the kinetic energies of 

v 2 V 2 V 2 

the different bodies, that is m — + m — - -f m — — -f- &c. 

2 2 2 

335. — Cor. If the bodies be the parts of a machine the ratio 
of their velocities to that of the driving point may be deter- 
mined from the mode of connection, e. g., if Vd is the velocity 
of the driving point and v r the velocity of one of the pieces 
and v r :vd:: n:l, then the equivalent weight at the driving point 

v 2 
is n 2 W and the total kinetic energv of the system is — 2 n 2 W. 

336. — Prop. The kinetic energy of a system of bodies equals the 
kinetic energy of the whole mass concentrated at the centre of gravity 
relatively to the external point and the sum of the kinetic energies of 
the bodies relatively to their common centre of gravity. 

When a system of bodies is in motion relatively to an ex- 
ternal point, each body has a velocity relative to the external 
point, the common centre of gravity of the system has a ve- 
locity relatively to the external point, and each body has ve- 
locity relative to the common centre of gravity. The velocity 
of each body relatively to the common centre of gravity is the 
resultant of the velocity of the body relatively to the external 
point and the velocity the body has in common with the com- 
mon centre of gravity. 

Let m n m 2 , m 3 , &c, be bodies of a system ; let v n v 2 , v 3 be 
their velocities relatively to an external point O, which is 
taken as the origin of rectangular axes ; let v be the velocity 
of the common centre of gravity in reference to the external 
point. 

Resolve these velocities into components parallel to three rec- 
tangular axes, namely, v lx , v iy , vi z , v 0x , v 0y , v 0z . Let v ; be the 
velocity of m, relatively to the common centre of gravity, then 
its components will be v' x , v' y , v' z , and v' x =vi x — v 0x ; v' y — Vi y 



136 

— v 0y ; v' z ^vi z — Voz ; or Vi x = v^+v'x ; v % = v 0y -f v' y ; v lz = 
v 0z -f-v / z . Then the kinetic energy of in, relatively to the ex- 
ternal point is 

yim x [(v 0x -fv / x ) 2 +(v 0y +v / y ) 2 - r -(v 0z +v , z ) 2 ] and similar expres- 
sions for the other bodies. The whole kinetic energy will be 
the sum of the expressions, or 

%I • m [(v^+vy+Cvty+yy+Cvte+v',) 9 ] = V 2 (v 2 9x +v 2 0y + 
v 2 0z ) 2- m + % Im (v x /2 + v y '*+v/ 2 ) + v 0x . - v -mv x ' + v 0y * 2mv' y 

V 0z - ■ fflv/. 

But lmv' x =0; ^-mv'y-0 ; I- mvV^°, (§ ) hence, the 
whole kinetic energy is 

% (v 2 ox+v 2 oy + v 2 0z ) Z-jn + rfS.m (v x ' 2 +v/ 2 -fv/ 2 ), the first 
term of which is the kinetic energy of the masses concentra- 
ted at the centre of gravity of the system and moving with the 
velocity of the centre of gravity relatively to the external 
point, and the second the sum of the kinetic energies 
of the bodies relatively to the common centre of gravity. 

337. — Prop. The kinetic energy of a rotating body about its. 
ce?itre of gravity equals its moment of inertia multiplied by the 
square of the angular velocity divided by 2Q. 

Let AB (Fig. 200) be a rigid body rotating around the axis 
C with angular velocity w. Let the linear velocity of a point 
Pi whose weight is W, and radius is r, be v,. Then v 1 =flTj. 

The kinetic energy of the point W will be 

v 2 W, co 2 a? 

W,- = -- wv = W,i\ a - = i, — 

2ff 2Q 2g 2ff 

Another point W 2 would have kinetic energy equal to 
Wo — = i 2 — 

iff ^g 

CO 2 CO CO 2 CO 2 co 2 

1, — + u— - +i 3 — + &c. = (i I +i,+i, + &c.) — ■= I— * 

*9 W 2ff 2ff 2ff 

V 2 

Compare with the kinetic energy due to linear velocity W — 

(§332)^ 



338. — Prop. The sum of the kinetic energies due to the rec- 
tangular compo7ients of the angular velocities about three axes 
of inertia is the same as the kinetic energy of the body. 

Resolve the angular velocity co around three axes of inertia 

into components co cos. «, co cos. /?, co cos. y. The three ki- 

co 2 cos. 2 a co 9 cos. 2 ft co 2 cos 2 y 
netic energies due to them are I. ; L ;I 3 

Their sum is I • — 

Compare this with kinetic energy due to the rectangular 
components of linear velocity. §333- 

.339-— Transformation of Energy. If a bullet is fired ver- 

2 

tically upwards work is performed (§309) equal to W — . 

It is obvious that when the ball has risen to a height at which 
v = all the kinetic energy has disappeared, but by virtue 
of its position the bullet has potential energy equal to m$s. 
Kinetic energy has thus been transformed into potential, just 
as the potential may be transformed into kinetic when the 
ball falls from its elevated position to the earth. This change 
is called the transformation of energy. The motion of a pen- 
dulum (Fig. 201) illustrates these changes. 

First imagine the pendulum to be in vacuo and frictionless. 

Let OB represent a pendulum. Suppose work mff ■ H,B 1 is 

performed on the bob B sufficient to lift it to the height Hx B^ 

1. At the point Bi the energy is all potential and equal to 

m^-BjHj. 2. At B 2 potential energy to the amount m^-Bihi 

v 2 
has disappeared, but kinetic energy m — = mg • B,h x has 

taken its place. 3. At B 3 all the potential energy has vanished 

but kinetic energy m — — m^-BjHi has appeared. 4. At 

2 <y 

B 4 the velocity has diminished and therefore kinetic energy, 
but potential energy m^B 4 H 4 is gained. 5. At B 5 the velo- 
city is zero, all kinetic energy is gone but potential energy 
equal to the original m^-B 4 H 4 appears. On the right hand 



■38 

side of B 3 potential energy has been transformed into kinetic 
and on the left hand side kinetic into potential. When the 
body falls from B t to B 3 its velocity is accelerated (§190) and 
the energy is said to be stored ; in moving from B s to B§ the 
motion is retarded and the energy is restored. In moving 
from B 3 to B 5 energy disappears in performing work against 
gravity which here acts as a resistance. 

Again imagine the pendulum in air and having friction at the axis 
of suspension. 

The original potential energy is now expended in produc- 
ing kinetic energy, and performing work against resistance,, 
hence at B 3 the kinetic energy is less by the amount of work 
performed in moving against the resistance of the air and 
friction. The sum of the kinetic energy and work equals 
however Wh. In moving up on the opposite side the energy 
restored is less, for the kinetic energy is less, and it still per- 
forms work against the resistance named. 

It is noticeable that in one-half of the motion of a pendu- 
lum gravity acts with the motion and in the other half against 
the motion and the two actions are equal. Such a force is a 
reciprocating force. 

Bending and releasing a bow, the to and fro motions of the 
piston of a steam engine, the vibrations of a tuning fork, the 
motions of a perfectly elastic body, all illustrate the same 
principle. 

340. — But it may be asked what happens when a cannon 
ball strikes a target ? Its velocity is destroyed and its kinetic 
energy has therefore disappeared. It has not however been 
annihilated, for the ball and the target where struck have 
both been warmed. Heat is a form of kinetic energy. Sound 
accompanies the stroke ; a flash of light appears ; an inden- 
tation is made. Light is partly kinetic and partly potential 
energy of wave motion in the luminiferous ether. Sound is 
partly kinetic, partly potential energy of wave motion in the 
air. The indentation is the result of work done. Energy also 
manifests itself in the electrical and magnetic states. 

Hirn proved that when a man ascends a flight of stairs he 



139 

loses a portion of his heat. He gains potential energy by his 
ascent. When he descends he loses his potential energy and 
gains heat. In his descent he converts a portion of his poten- 
tial energy into kinetic energy and then arrests his motion 
and converts the kinetic energy into heat. 

The decomposition of water by electrolysis is an instance 
of the potential energy of chemical separation effected by 
chemical work. The hydrogen and oxygen molecules, fall 
towards each other, when a spark is applied, and the energies 
of heat, light and sound appear. In the gas engine some of this 
energy is applied to produce motion of the piston. 

Many other illustrations, combustion, plant growth, the 
sun's heat, etc., will suggest themselves. 

341. — Availability of Energy, Dissipation of Energy.— 

When a brake is applied to a car-wheel, the brake and the 
wheel are both heated. The heat produced is the equivalent 
of the work performed against friction. The heat of the 
wheel and shoe radiates to other bodies and becomes unavail- 
able for doing work. In every transformation some energy 
is converted into heat, which becomes diffused and useless for 
purposes of work. As this operation is going on continually 
the energy of the universe is running down and matter will 
finally assume a uniform temperature. Part of the energy is 
available for man's work and part is nonavailable. The avail- 
able energy tends to zero ; the nonavailable tends to a maxi- 
mum. 

342. — Conservation of Energy. Newton's second law of 
motion (§263) shows that F^ma. Imagine the point of applica- 
tion of the force to be displaced through a small distance s 
and multiply both members of the above equation by s, then 

v 2 

Fs = mas, but v = y 7 2as (§198) or v 2 = 2as and a = ■ — 



2S 



T- V V TTT V~ 

Fs = m — Sr=m — = W — 



2S 2 20 

Now the first member of this equation is what is called 

(§327) potential energy, and the last (§332) kinetic energy, 

hence, this expression obviously gives another method of 



140 

measuring an unbalanced force — by the change of kinetic energy 
it produces in acting over a given distance instead of by the 
change of momentum it produces while acting for a given time. 

§263. 

343. — If m be urged by a force along a straight line and the 
velocity changes from v, to v 2 the potential energy exerted on 

, , , -, • mv 2 2 mv, 2 , , • , 

tne body is— — or the potential energy exerted is equal 

to the increase of kinetic energy. If the force is a resistance 
the decrease of kinetic energy is equal to the work performed. 

It is from considerations of this kind and also from nu- 
merous experiments and deductions that the principle called the 
Conservation of Energy has been established. It is usually 
expressed as follows : 

The sum of the potential and kinetic energies in the universe 
is a constant quantity. 

It may also be expressed in terms of work, as follows : 

In any interval of time of a body's motion the potential 
energy exerted plus the energy restored is equal to the work 
performed plus the energy stored. §§339, 340. 

Cor. If the motion of a body is uniform it is obvious that the 
potential energy exerted is equal to the work performed, or Fs = 
F r s r . 

In the application of this principle to machines the principle 
is often stated that the work done on the machine equals the tvork 
done by the machine. 

344. — Illustrations of the Principle of the Conservation 
of Energy in Uniform Motion. 

If the principle is admitted the law of balance for machines 
may be directly deduced. 

Let the bent lever (Fig. 202) be moved through a very small 
distance so that F, describes A/a and F 2 describes A/b. Then 
F, • A/a = F 2 ■ A/b or F, :F 2 ::A/b: A/a :: CP>: CP 2 but this 
is the law of balance of the bent lever. §131. 

Let the wheel and axle (Fig. 203) be turned through a small 



141 

angle. Then F, ■ P^n = F, ■ P 2 n or F, : F 2 :: P 2 n: P T m :: CP 2 : 
CP 3 . §140. 

In the fixed pulley (Fig. 97) V, ■ s T = F 2 • s,, but s, =s„ 
hence F x = F 2 . §142. 

In the shigle movable pulley (Fig. 98) let F 2 be moved over 
the small distance x. Then, if there are n strands in the rope 
F x will be displaced a distance nx. The principle of the con- 
servation then gives F t ■ nx = F 2 ■ x or F, : F 2 :: x: nx: l : n or 

F=-- 2 . §142. 
n 

L,et the screw (Fig. 100) have the power displaced! 2~r, then 
the weight will move over a distance equal to the pitch p. 
Then F 1 : F 2 :: d : 2^r. §144. 

In the inclined plane (Fig. 204) F x • ffp = F 2 ■ g i p l or F, : F 2 
nffipi : ffp :: gg x sin. ffig^x : gg x cos. FEB : sin. ^i#pi : cos. FEB 
::sin. : sin. F#N. §143. 

In the compound pulley (Fig. 1 1 1) let the lower pulley be 
lifted a distance x ; the next will be raised 2X ; the top one 
2 ■ 2x = 2 2 x ; the n th one will be raised 2 n x ; hence Fi • 2 n x — 
F 2 ■ x or Fi: F 2 ::x: 2 n x::l: 2 n . §158. 

In the compound pulley (Fig. 112) if the weight be raised a 
distance equal to x, the first movable pulley at the top will 
descend a distance equal to x ; the lower pulley will descend 
(2+i)x ; if there were another it would descend 2(2+i)x 
+x or (2 2 +2-f-l)x. If there were n pulleys the power would 

descend (2 n_1 + 2 n ~ 2 + -f-l)x==(a n --l)x. Hence Fi . (2 11 — l)x 

=F 2 • x or Fi : F 2 :: x : (2 n — l)x :: 1 : 2 n — 1. §160. 

In the pulley (Fig. gSa) F 1 • nx=F g F 2 • x. Fi : F 2 :: x : nx :: 
1 : n. §142. 

Analogous proofs can be given for other machines. 

In all the preceding cases the displacements are supposed 
to take place in the same time or the parts to move uniformly. 
Hence the F multiplied by tv must equal F r multiplied by t^. 
But t and ti are equal, hence the proposition is sometimes 
called the "principle of virtual velocities" and expressed as 
follows : The power multiplied by its velocity is equal to the weight 
multiplied by its velocity, which is only another way of saying, 
u what is gained in power is lost in speed. 11 



142 



THE THEORY OF POTENTIAL 



345.— Field of Force. Definitions. Afield of force is a re- 
gion in which work is performed in moving a mass from point 
to point. When the body is released kinetic energy is pro- 
duced. §339. In all that follows a conservative system is 
considered in which the work performed is equal to the kine- 
tic energy produced. §327. Such a field surrounds the earth, 
electrified bodies and magnets. 

346. — Lines of Force. A field of force is supposed to be 
mapped out with lines, tangents to which represent the direc- 
tion of the resultant force at every point in the field. These 
lines are called lines of force. In the earth's field the plumb- 
line places itself along these directions ; in the field of an 
electrified body a charged body will move along them ; in the 
neighborhood of a magnet a freely suspended magnetic needle 
will place itself tangent to them. " Stream lines," " lines of 
flow" of heat are other illustrations. 

The. positive direction of a line of force is the direction along 
which motion takes place when a free body is placed in a field 
of force. A free mass of ordinary matter will move towards 
the centre of the earth when placed in the earth's field ; the 
direction a body charged with a plus unit of electricity in an 
electrical field and the direction an isolated north pole would 
be urged in a magnetic field are positive. The direction of 
"stream lines'' is the direction of flow of the liquid from 
higher to lower level. The lines of flow of heat are directed 
from places of higher to places of lower temperature. The 
negative direction is opposite to the positive. 

347. — It is obvious that two lines of force cannot intersect, for 
there cannot be two directions of resultant force. 

348. — Strength of Field at a Point (or the resultant force 
at a point, the intensity of force at a point, the intensity of 



143 

the field at a point) is the force, attracting or repelling, ex- 
erted by the body or bodies producing the field on unit mass 
placed at that point. 

The unit mass selected for the earth's field is one gramme, for 
an electric field a small body charged with a plus unit of elec- 
tricity and for a magnetic field a unit north pole. 

Since the intensity of a force is measured by the change of 
momentum it will produce in a given mass (§261) F — ma, it 
is obvious that the strength of the field is the same as the ac- 
celeration, F = 1 X a, ox the force per unit mass. Hence in the 
earth's field in the latitude of Paris, at the level of the sea the 
strength of field is 981 dynes. §264. 

349. — A field of force is known when, (1) the direction of the 
force at every point, and (2) the strength of 'the field are given. 

350. — The law of attraction which applies to gravitation, 
electricity and magnetism, is that the force varies as the pro- 
duct of the masses of the bodies and inversely as the square of 
the distance between them. §269. 

The following results which may be easily obtained by ana- 
lytical methods, are of special interest in the sciences of elec- 
tricity and magnetism : 

1. The strength of field at a point due to a uniform straight 
line equals a constant multiplied by the product of the area of 
the cross-section of the line, its density and the sine of half 
the angle included between two lines drawn from the extrem- 
ities of the given line to the point, divided by the perpendic- 
ular distance drawn from the point to the line and its direc- 
tion bisects the same angle. 

2. The strength of field at the centre of a circular arc due 
to the attraction of an arc of the circle is the same in magni- 
tude and direction as that of a line which touches the arc and 
is limited by lines drawn from the centre to the extremities 
of the arc, both having the same density, and area of cross 
section or the attraction of the arc is the same as a mass equal 
to the chord, with the same density as the arc, concentrated 
at the end of the radius which bisects the arc. 



144 

3. The strength of field at a point, on a line perpendicular 
to a uniform circular plate at its centre, equals 20r/?t (1 — cos.y) 
in which C is a constant, p the density of the plate, t its thick- 
ness and y half the angle subtended by the plate as seen from 
the point. If the particle is on the plate the attraction be- 
comes 2~/?t. If the particle is in the interior of the plate at a 
distance tj from one surface and t 3 from the other the attrac- 
tion is 20y? (t 2 — t,). In passing from one side of an indefinite 
attracting plane to the other the attraction changes from 
+ 20:/? to — 20y?, and the difference of attraction between 
the two sides is 4O;/?. 

4. The attraction of two equal parallel disks on a particle 
situated on a line perpendicular to both of their centres — one 

disk attracting and the other repelling is 2x (? 

^(x+df+r,- 

x 1 . 

T^^^-i , in which r t is the radius of each disk, d the dis- 

^x.+r- J 

tance between the two disks, x the distance of the point from 

the nearest disk and <* the superficial density of either plate. 

5. The attraction of a right circular cylinder on a particle 

placed on its axis is 2 Or/? [ 1 — i/(h— I) 2 + r, 2 + -j/h 2 -f- r : 2 ] in 
which 1 is the length of the cylinder, h is the distance of 
the particle from the upper base and r, is the radius of the cyl- 
inder. 

6. The attraction of a uniform right circular cone on unit 
mass placed at its vertex is 2 Or/? (1— cos. y) h. 

7. The attraction of a uniform spherical shell of very small 
thickness upon a point exterior to it is the same as if its mass 
were concentrated at its centre. §270. If the particle is 
within the shell the resultant attraction is zero. §271. If 
the particle is just outside the attraction is 4O7?. The at- 
traction of a hollow sphere on an element of its surface is 2/T 6 2 s 
in which s represents the elemental surface and a the surface 
density of the shell. The attraction of a hemispherical shell 
upon a particle at its centre is zQnp and hence is independent 
of the radius. 



145 

8. The attraction of a homogeneous sphere upon a particle 
exterior to it is the same as if all its matter were concentrated 
at its centre. If the particle is just outside the sphere the at- 
traction is - Qnpv. 

9. The attraction of a solid homogeneous sphere upon a par- 
ticle within it is proportional to its distance from the centre. 
§272. 

351. — Potential. The direction of the force and the strength 
of the field may be expressed in terms of potential. 

352. — Mutual Potential Energy. Definition. The mutual 
potential energy between two bodies is the amount of work which 
can be obtained from their mutual repulsion when they are allowed 
to separate to an infinite distance. 

353. — Prop. If the law of repulsion is that the force equals 
2 the mutual potential energy, or the work performed 



s s 



in the separation of two masses from distance s to infinity, is 
equal to a constant multiplied by the quotient arising from di- 
viding the product of the two masses by the distance between 
them, for 

J F I ds=/f^?) ds=Cmjn 2 J ids^Ci^mJ^ s ds= 

Cm.m.f-I +_L) = C^*. 
\-°° s / s 

1. If the separation is from distance s, to s 2 the potential 

energy becomes Cmm 2 ( — — — ) . Hence, if a force, which 

> \s 1 sj 

repels the bodies to twice a given distance, does a certain 
amount of work, to repel them to infinity it would perform 

only twice the work, for work = Cn^m,, ( — — ] = Cm 1 m 2 

\ S 2S/ 

— ) = l — - , which is half the quantity given in the prop- 



\ 2S / 2 

osition. 



2. When two repelling (+) bodies are in contact their 

, .. , . . r Cmm 
potential energy is a maximum, for energy = — = 00. 



146 

3. When two repelling bodies are separated a finite distance 
their potential energy always has value. 

4. If the force between the bodies is an attraction ( — ) (§327), 
when the bodies are in contact their potential energy is nega- 
tive and numerically a maximum. §327. 

6. When the attracting bodies are separated to infinity the 
mutual potential energy is negative zero. 

354 — Absolute Potential. Definition. If the mass of one 
of the bodies is taken as unity then the mutual potential energy be- 
tween them is called the absolute potential of the point where unit 
mass is placed due to the attracting or repelling body. 

It follows, therefore, that a field of force may be explored 
by imagining unit mass placed at each point in the field and 
determining the amount of work performed by its motion from 
the point to an infinite distance. If the field is electrical the 
unit employed is one of positive electricity ; if the field is 
magnetic a unit north pole is used ; if the field is the earth's, 
a gramme. §348. 

355. — Direction of Motion. If the force is repelling, unit 
mass will move to infinite distance where the repulsion will 
cease; it will move from places of higher to places of lower po- 
tential. §327. In this case the force will do work. If unit 
mass be moved from a place of lower to a place of higher po- 
tential work will be performed upon it against the force of re- 
pulsion. 

In the case of gravity, since it is attractive, potential 
will be negative ; a point on the earth's surface has minus po- 
tential ; at an infinite distance the potential is zero. For 
convenience a point on the earth's surface is regarded as hav- 
ing zero potential. A body when in contact with the earth 
has no potential energy. §327. 

356. — Difference of Potential. It follows that the differ- 
ence of potential between two points is the amount of work 
performed in moving unit mass — a plus unit of electricity or 
unit north pole of magnet or a gramme from one point to the 
other, for 



H7 

Let Y Al = the absolute potential at A„ that is the amount 
of work necessary to move unit mass from A, to an infinite 
distance beyond the influence of the attracting or repelling 
mass, and Va, = the absolute potenial at A 2 , that is the amount 
of work necessary to move unit mass from A 2 to infinity, then 
Va, — Va 2 = the difference of potential between A T and A 2 or 
the amount of work necessary to move unit mass from A x to 
A 2 , and if F — the resolved part of the force acting between 
two near points A l and A 2 and A,A, the distance between 
them, then (§308) 

F • A t A, = V A — Va, and 
y A Y\ 

F= — ~ — 2 ; that is, the average force is the average rate 

AjA Q 

at which potential energy changes along Aj A 2 , or it is the rate 
of change of potential per unit length. 

Experiment shows that there is no electric force inside a 
charged conductor, hence there is no difference of potential, 
hence potential on the interior is constant, hence the surface 
of a closed conductor is equipotential. §362. If this were not 
so there would be a flow of electricity from the places of higher 
to places of lower potential. 

Cor. If AjA 2 is very small the force is in that direction the 
rate of change of potential at that point in a given direction. 

357. — Datum Line. The point of absolute potential is at 
an infinite distance from all repelling bodies, but the position 
of this point is immaterial from the fact that difference of po- 
tential is alone considered. The base line may be taken any- 
where and called the place of zero potential. If the unit mass 
has work performed on it to move it from this place to an- 
other, the second place has positive potential ; if the force 
does work in moving unit mass from this place to another the 
second has negative potential. 

" Low water mark" is a similar arbitrary point used as a 
standard of reference. Any point above it is at a higher level, 
any point below it is at a lower level. The sea-level is gen- 
erally employed as a standard. No difficulty would be ex- 



148 

perienced if a point were spoken of as having a certain num- 
ber of units of plus or minus potential energy instead of being 
so many feet above or below sea level. 

The difference of potential for two places near the sea level 
may be expressed in foot-pounds or centimetre grammes. 
§312. 

Fs = Ws = m^s = igs = 32. 2h in which h = the difference in 
height above the sea-level in feet or Fs = Ws = mg$ =l-g-s= 
981I1 centimetre-grammes. 

Difference of potential produces a flow of electricity just as 
difference of level produces a flow of liquid and difference of 
temperature a transference of heat. 

Since difference of potential produces a flow of electricity it 
is called electro-motive force ; difference of level produces a 
flow of liquid by producing pressure or hydromotive force 
which causes the flow. 

In electricity the plus unit is repelled along the lines of 
force into neighborhoods of less, zero or minus potential. The 
body must always pass through the point of zero potential to 
get from places of positive to places of negative potential. 

358. — The following consequences proceed immediately 
from the definition of potential. 

Since potential is the mutual potential energy between a 
body and unit mass placed anywhere in its field the proposi- 
tions under energy (§327) apply. 

1 . — The difference of potential between A, and A 2 is inde- 
pendent of the path followed by the unit mass in moving from 

A, to A 2 . 

2. — The potential changes most rapidly along the line of 
force ; for, if the force has any other direction since it can be 
resolved into components, the rate of change would be less. 
The inclined plane furnishes an illustration of this truth. 
At right angles to the line of force the change of potential is 
nothing. 

3. If the difference of potential equals unity, Va, — Va 2 = 1, 



H9 
and the force varies inversely as the distance between the points 



A, A, 

4. If the distance between A, and A 2 is unity then the force 
is measured by the change of potential. 

5. In free space the value of potential never changes sud- 
denly, for if the difference of potential has value when t]ie dis- 
tance between A l and A 2 is zero, the force will be infinitely 
great, which never occurs. §357. 

359. — Another Form of Absolute Potential. Prop. The 
absolute potential at a point or the work obtained from repul- 
sion in driving unit mass from the point to infinity is — in 

s 

which m is the mass of the particle repelling unit mass and s 
is the distance between the two points. §328. 

Cor. If there are several points at which there are masses 
mi, m„ m„ &c, then the absolute potential of a point due to 

m is— 1 -, due torn., — 2 , due'tom 3 , — 3 , and the whole potential 
1 s, s a s 3 

will be — ■ -f — - 2 -f — - = I — , that is, the potential at a point is the 

Sj So S 3 s 

sum of the quotients obtained by dividing each of the acting 
parts by its distance from the given point or the amount of work 
necessary to move unit mass from the point to infinity equals 
m 



V 

S 



360. — Another Form of Difference of Potential. It has 



just been shown that the absolute potential of one point is 



111 



s. 



(within a constant). It can also be shown that the absolute 

potential of another point is — . Hence, the difference of po- 
rn m / 1 1 
tential between the tw 7 o points is = m 

Sj s 2 \ s x s 2 

that is, the difference of potential between two points is equal 
to the product of the mass of the repelling body and the differ- 



150 

ence of the reciprocals of the distances of the two points from 
the body. 

361. — Cor. If there are other particles in the field the sum 
of the works done by the attraction of all the particles while 
unit mass is moved from one point to another is, [if m„ m 2 , 
m 3 , &c, are the attracting masses, the unit mass being moved 
from A, to AJ 

1 1 \ . ~ /I 1 \ . ~ /l 1 



Cm 
or. 



+Cm * ivrvj +Cm3 (s-?-s7')^ vm (H) 



s, s' t s/7 U s'f s/'/ ^ s 2 " s/ 

s.„ s/, s/' are the distances of the final position of A 2 and 
s n s/, s/ ; the distances of the initial position of A, from the 
points where m„ m 2 , m 3 , &c. are, that is, the whole work done 
is equal to the sum of the works done against each element of 
the system. 

Cor. If A, is at an infinite distance from every attracting 

point — = — : =0, and the work = C — ~f — -f — -f&c. ) 
s, s/ \ s, s 2 s 3 / 

= the absolute potential. 

Cor. If A. is at a great distance - — =0 and Va, == w - — 

s 2 s 

The preceding considerations show that potential is a scalar 
quantity and is also arithmetically additive. 

362. — Equipotential Surfaces. It is evident that surround- 
ing a body there are points so situated that their potential is 
the same. A surface passing through such points is called 
an equipotential surface. 

Properties of Equipotential Surfaces. 

1. An equipotential surface must cut all the lines of force 
at right angles. In the case of gravity a level surface is an 
equipotential surface, every part of which is perpendicular to 
the direction of the plumb-line which represents the direction 
of resultant force. 






i5i 

2. Different equipotential surfaces cannot intersect because 
they are perpendicular to the lines of force. 

3. No work is performed in moving matter along an equi- 
potential surface, as is illustrated in the case of the floors of 
buildings, in electricity and magnetism. 

4. The work performed in moving from one surface to an- 
other is independent of the path described. 

5. The work performed in moving from any point on one 
surface to any point on another is constant. 

6. The rate of change of potential along an equipotential 
surface is zero. 

7. The equation of an equipotential surface is V = C, be- 
cause the force resolved along an equipotential surface has no 

value. The force resolved along a tangent to a curve s is — 

as 

but the force will be perpendicular to the arc if — - = 0, in which 
case V = constant. 

8. If n is the normal to an equipotential surface the whole 

force F = — and acts in the direction V increases, 
dn 

363. — A System of Equipotential Surface. 1. Equipoten- 
tial surfaces may be so selected that unit work will be per- 
formed in moving from one to another. It is evident that 
the distances between them must increase as the distance from 
the body producing the field increases, for since, Va, — Va 2 = 

F • A X A 2 = 1, F= , that is, the force at any point varies 

inversely as the perpendicular distance between two successive 
surfaces near that point. 

2. The distance between successive concentric equipoten- 
tial surfaces is proportional to the square of their mean dis- 
tance from the centre of the mass producing the field. 

Let V = the potential of a spherical equipotential surface 







152 










whose radius is r,. 


V+l — the same of a 


surface whose radius 


is r 2 . 














Then V= 


=C™; 


v+i=c5. 

r 2 










u 

A 2 


= Cmi 


- jr ' = Cm v+i 


• 










= Cm 


1 Cm ; = 

V V+i 


- Cm 


l l - 


1 ) 


— Cm 


*V ^1 


rv 


V+i/ 




/V+i-V\ 
|V(V+i)/ 


= Cm 


(■ J Ucm 


1 




r>r 2 
Cm * 




Cm 


Cm = 





r r 

3. If the spherical equipotential surfaces are taken #/ equal 
distances from one another, the work required to move mass 
from one to another varies inversely as the square of the 
mean distance of the two surfaces from the body producing 
the field. 

Let V T and V 2 be the potentials of two concentric spherical 
equipotential surfaces and r and r-f l be their radii, 

then V, = Cm - ; V 2 = Cm ~ ;, V, - V, - Cm ^pyy 

364. — Geometrical Principles. Definitions. A conical 
surface is generated by moving a straight line about a point 
fixed in the line. 

The Solid Angle of the Cone. From the vertex of a cone 

as centre describe any number of spheres. The segments of the 

concentric spherical surfaces, thus cut by the cone will be 

similar and their areas will be proportional to the squares of 

the radii. Let S, (Fig. 205) be a surface which subtends the solid 

angle of the cone at unit's distance from the vertex, O, and let 

S2 be any concentric surface of radius r 2 . Then S ]; : S 2 :: l 2 : r/ 

S 
or S, = — ? . This ratio is taken as the numerical mea- 

r 2 

sure of the solid angle of the cone just as is taken as 

radius 

the corresponding measure of a plane angle. 



153 

365. — The following propositions are obvious : 

1. The sum of all the solid angles about a point equals 4-. 
For the area of a spherical surface described about the point 

A.7IX 2 

equals 4~r 2 , hence the solid angle = — — = 4-, which is the 

whole area of a sphere of radius unity described about the 
point. 

2. The sum of the solid angles of all the conical surfaces 
which can be generated without intersecting each other is equal 
to 2~, because the solid angles of vertical cones are equal. 

3. TJie solid angle subtended by a?iy closed plane curve at a 
point which is in the plane of the curve and within the curve 
equals 2~. 

Let Fig. 206 be in the plane of the paper and O be within 
the curve and in the plane of the paper. Then radiating lines 
intersect a spherical surface in a great circle of the sphere. 

4. The sum of all the solid angles subtended by the elemental 
areas of any surface at a point outside of the surface equals zero. 

Let O (Fig. 207) be the point without the closed surface S. 
It is obvious that all the elements subtend the same solid an- 
gle but the signs are different at the different elements. At 
A, the outside of the surface is turned towards O ; at A 2 the 
inside ; at A 3 the outside and so on. Hence, the sum of the 
solid angles subtended at O by the elements is (A) 4- ( — A) 
-f (A) -f- ( — A) which equals zero. 

An orthogonal section of a cone is made by a spherical sur- 
face described from the vertex of the cone. 

An oblique section is a section inclined to the orthogonal 
section. 

5. The area of an orthogonal section is equal to the area 
of the oblique section multiplied by the cosine of the angle of ob- 
liquity of the section. 

In Fig. 208, let S^ be an orthogonal section and SS an 
oblique section and the angle of obliquity then SS = S,S, 
cos. 0. 



*54 

6. Cor. The area of an oblique section of a small cone equals the 
product of the distance of the section from the vertex of the cone and 
the solid angle divided by the cosine of obliquity of the section. 

S S = M. = OS> A since a =^4' 

cos.6> cos# 8 ' r, 

Flow of Force, (number of lines of force, surface integral 
of normal force-intensity, normal flux of force through the 

surface outwards, j F n dS.) 

366. Definitions. The normal force acting on an equipoten- 
tial surface multiplied by the differential of the surface on 
which it acts is called "flow of force.' 1 '' 

The volume of a liquid passing through an area dS is equal 
to the velocity multiplied by the differential of the area or vdS. 
The above expression is called "flow of force" from its anal- 
ogy to the flow of a liquid. 

The flow of force per unit area equals the normal compo- 
nent of the force. 

367. — Prop. If a particle (dm) is situated without a closed sur- 
face drawn near it, provided the attraction is according to the law of 
inverse squares, the il surface-integral of the normal force intensity" 
is equal to zero ; if the particle is within the surface, the surface-in- 
tegral is equal to — 4^m. 

1. Let the point be outside the surface. 

Let O (Fig. 209) be the particle and S the surface. Draw 
the line 00! cutting the surface at points A„ A 2 , A 3 , A 4 . De- 
scribe about O a conical surface enclosing a solid angle dA, 
and on the indefinitely small sections made by the cone draw 

dA 
perpendiculars outwards. The force intensity at A, is C . 

The component of this force resolved along the normal drawn 

dA 

outwards A,N, = C cos. OA,Ni. The element of the sur- 

OA, 2 

face integral = C-^— cos, OA>N, dS. But _gg ^ s - QA » N . 
OA t 2 l l OA, 2 



155 

= dA. Therefore the element of the surface integral — 
CdmdA, and the surface integral = C dm ( dA=o. §365,4. 
That is, if F n represent the magnitude of the normal compo- 
nent on any element dS, I F n dS==C dm j dA = o. 
2. Let the point be within the surface. 
fF n dS=-Cdm JdA = _ Gtfrdin. §365,1. 

368. — The expression in terms of potential becomes j F n dS— 



/ 



dV 

-— dS— o or— ±710. 

dn 



369. — LaPlace's and Poisson's Equations. At any point 
O (Fig. 2 10) take the indefinitely small volume dxdydz. LetX, 
Y, Z be the resolved parts of the force along the axes Ox,Oy,Oz. 
Then the ' 'flow of force' ' through the face Oabc= — Xdzdy = — 

dV dV 

— . dz dy, and the flow through the opposite face is = — dz dy 
dx dx 

-4- — I — dy dz dx = — - dz dy + - — dx dy dz, hence, the 
dx \dx / dx .~ dx 3 

dV 

total flow parallel to Oz through the element = dz dy 4- 

dx 

dV d 2 V d 2 V 

— dz dy + - — dx dy dz = — — dx dy dz. Similar expressions 

dx y dx 2 y dx 2 y * 

may be found for the other faces and the whole flow becomes 

/d 2 V d 2 V d 2 V\ 

/a_v + oy + o_v dx d dz = v ( dx d dz . Jf there 

\dx 2 dy 2 ^ dz 2 / J v y ' 

is no attracting mass within the element, p 2 V (dx dy dz)=o. 

Hence r 2 V = o. Hence at every point in a space where there 

is no attracting mass r 2 V = o. This is LaPlace's equation. 

370. — Again, ifO is within the attracting mass, r 2 V= — Cprp 
if p is the density at O. This is called Poisson's equation. 

371. — Tubes of Force. Definitions. A tube of force is a 
surface bounded laterally by lines of force, and at the ends by 
indefinitely small equipotential surfaces. 



156 

Illustrations. — i. If the attracting body is symmetrical 
about a point the lines of force radiate from the point and the 
tubes of force are conical. 

2. If the attracting body is arranged symmetrically in in- 
definitely long cylindrical shells around an axis, the lines of 
force are normal to the axis and the tubes of force are 
wedges. Fig. 211. 

3. If the attracting mass is arranged in infinite parallel 
planes of equal density, the lines are normal to the planes and 
the tubes are cylinders. 

372. — Property of Tubes of Force. Prop. If the tube docs 
not cut through attracting matter the whole force is at every point 
tangential to the direction of the tube and inversely as its normal 
transverse section. 

For, let Fig. 212 represent a tube of force ; then by Gauss's 
theorem (§367), the surface integral of the normal force inten- 
sity or the " flow of force'' equals o. 

JV ni dS-JV n2 dS=o. 

JXdS =jF n2 dS 2 or F nt S = F n2 S 2 . 

1. If the tubes of force (Fig. 213) are about a point (§371, 1) 
then F ni : F 1>2 :: S, : S a , but S t : S, :: D 2 2 : Df . • .F n3 : F„ 2 :: D 2 2 : D,* 
that is, the force varies inversely as the square of the distance from 
the vertex of the conical tube. 

2. In tubes of cylindrical shells (§370, 2) F n ,: F n2 :: S,: S 2 , but 
S, : S 2 : : r 2 : r 2 , . * . F ni : F n2 :: r 2 : r„ that is, the force varies inversely 
as the distance from, the axis. 

3. If the mass is arranged in infinite parallel planes (§371, 3) 
F n , : Fn.rr.S,: S„ but S, iS.irS, : S 2 , .*. F n = F n , that is, the 
force is constant at every section. 

373. — Prop. If the tube cuts attracting matter j F n dS — 4~m y 

F^dSj — F n2 dS 2 = 47ral/>, in which 1 is an indefinitely small 
length, a the area of cross-section of the tube and p the den- 
sity of the matter. 



157 



DYNAMICS OF LIQUIDS AND GASES. 



374. — Statics of Liquids and Gases.— Definitions. Liquids 
and gases are both characterized by great freedom of motion 
among their molecules. Liquids and gases are not perfect 
Viscosity exists to a greater or less extent in all of them. In 
what follows they are, however, regarded as having perfect 
mobility. On account of this property liquids tend to retain 
their volume but not their shape. Gases tend to retain neither 
their shape nor volume. Liquids are practically incompressi- 
ble. Gases are very compressible. 

375. — 1. Pascal's Principle. It follows, therefore, that if 
forces are applied to weightless liquids or gases, unless the 
forces are equally distributed over their interior surfaces, 
change of shape must result, or if a liquid or gas is in a con- 
dition of equilibrium every particle must be pressed equally 
in all directions. 

376. — 2. Principle of Transmissibility of Fluid Pressure. 

It also follows that whatever pressure foreign or due to its own 
weight is communicated to the external surface of a liquid or 
gas, will be transmitted to all parts of the liquid or gas with- 
out change. 

Let AB (Fig. 214) be a cylindrical vessel of small height 
standing upon its base. In the sides are placed pistons A,, 
A 3 , A 8 , etc., and the whole is filled with water. The weight 
of the water acting equally upon all the pistons may be thrown 
out of consideration. If a force F Pl is applied to the piston A, 
it will be transmitted without change to all the other pistons. 
If the area of A, is the same as each of the others each will be 
pushed out with a force equal to that with which A x is pressed 
in. Imagine now the whole surface of the cylinder to be made 
up of movable parts the size of A n the bursting pressure will 
then equal the pressure on A x , the unit of surface, multiplied 



'58 

by the whole interior area S or calling p the intensity of pres- 
sure the bursting pressure R p = pS ; or if F Pl is distributed 
over the piston A a the area of which is S x and the total interior 
area of the vessel is S 2 , then, if R p is the bursting pressure 

F p , :R p ::S,:S s or R p =|iF Pl . 

377- — 3- The Principle of Perpendicularity of Pressure. 
It is obvious, that whatever pressure is communicated to a 
liquid or gas will be normal to the surface pressed upon; for 
if the force is oblique it may be resolved into two components, 
one normal to the surface, the other tangential, the latter of 
which will produce motion, which is contrary to the suppo- 
sition of equilibrium. This principle maybe roughly illus- 
trated by the apparatus represented in Fig. 215, which consists 
of a sphere connected to a pump. The. surface of the sphere 
is perforated with small holes. When the sphere is filled with 
water and the piston of the pump forced down jets issue from 
the holes which are perpendicular to the sphere at each point. 

378. — The principle of the transm^ssibility of fluid pressure 
is obviously only an illustration of the principle of the conser- 
vation of energy, for if a pressure F Pl acting upon a piston 
whose area of cross-section is S, drives out pistons whose com- 
bined areas are S, with a force R p , then F Pl : R p :: S 3 : S 2 , but 
as the same quantity of liquid is driven from the small cylin- 
der whose piston is A, to the large enclosure whose pistons 
are A 2 , A 3 , A 4 , &c, the distance the first piston moves, s n 
must be as much greater than the distance the second piston 
moves, s,, as the area of the first is less than the area of the 
second or s 7 : s 2 :: S 2 : S„ hence F Pl ■ s x = R p • s 2 , which is the 
principle of the conservation of energy in uniform motion. 
§343. Cor. 

379. — The hydrostatic press, invented by Bramah r in 1796, 
is a practical application of this principle. It consists (Fig\ 
216) of two cylinders B n B 2 of unequal areas of cross-section 
S n S 2 , connected by means of a small pipe, the whole being 
filled with a liquid. Two pistons, A n A 2 , are placed one in 
each cylinder, free to move. If the small piston is forced 



1 59 

down the large one rises with a force which is as much 
greater as the area S 2 exceeds S r This principle is applied 
in testing the strength of pipes, boilers and cannon. 

The hydrostatic press is employed in many operations re- 
quiring the exercise of enormous pressure, as in pressing cot- 
ton, hay, books, cloth, paper, extracting the juice from plants, 
raising great weights, testing the strength of cables, &c. . If 
a great force is applied to the large piston a comparatively 
small one will be exerted at the other. 

380. — Illustrations of the same principles may be found in 
the blowing of soap bubbles, in the operations of glass blow- 
ing, &c. Fig. 217 represents a sphere, in different parts of 
which are placed siphon gauges, 1, 2, 3, 4, containing mer- 
cury. When the gum tube G is blown into, the mercury in 
the different gauges is displaced, showing that the pressure 
is transmitted in all directions. 

381. — The preceding principles maybe applied to the weight 
of the liquid or gas itself. The pressure due to the weight of 
a liquid or gas is transmitted without change to all parts of the 
liquid or gas and containing vessel and is normal to the surface 
pressed upon. If an additional external force is applied to the 
top of a liquid or gas the total pressure equals the sum of it 
and the pressure due to the weight. 

382. — Prop. The total pressure upon any surface in a liquid 
or of the containing vessel is equal to the weight of a column 
of liquid whose height is the distance from the free surface 
of the water to the centre of gravity of the surface pressed 
upon and whose base is the given area. 

Let the surface S of the vessel (Fig. 218) containing a liquid 
be divided into an infinite number of very small parts dS. 
Calculate the pressure on each element and add these pres- 
sures. The pressure on an element may be determined as fol- 
lows : Imagine the liquid in the vessel to be made up of an in- 
finite number of vertical prisms the height of each one being 
z and the base dS. Each prism may be further imagined to 
be cut by horizontal planes into an infinite number of flat 



i6o 

plates. It is obvious that the pressure on the lower base of 
the upper flat plate of any prism is the weight of the liquid 
above it ; that the pressure on the lower base of the second 
plate is its own weight plus the weight of the first and on the 
third its own weight plus the weight of the first and second. 
Hence, the pressure on the bottom of the original prism will 
be the weight of the column whose height is the distance of 
the base below the surface of the liquid and the base the ele- 
mental area. If dS is the lower base of the prism, z its height 
and zv the specific weight of the liquid, zdS will be the volume 
of the prism and zz^dS will be its weight, or the pressure on 
the base will be zz£/dS. The same thing is true if the lower 
surface of the element is inclined to the horizontal. The sum 
of these elemental pressures may be represented by - ■ z z#dS. 
But this expression is the sum of the moments of the compo- 
nent weights which by §H5# is equal to the moment of the 
resultant weight or z Zw* dS. Hence, the total pressure R p = 
~ • zz^dS = z Iw&S = z-'W-2- dS = hz£>S. w is placed outside 
of 2 because the liquid is supposed to be homogeneous. 

383. — Pressure independent of the shape of the vessel and 
the quantity of the liquid. 

The equation R p = hz£/S shows that the pressure varies as 
the surface pressed upon, the depth of its centre of gravity be- 
low the level of the liquid and the specific weight of the liquid 
and upon no other conditions. The pressure exerted on the 
bottom of a vessel may therefore be greater or less than the 
weight of the contained liquid. If vessels of different shapes 
and capacities (Fig. 219) with the same area of bottom, have 
a well-fitting brass disk pressed against them by a counterpoise 
at the opposite extremity of the beam of a balance it will 
be found that the disk will be detached by water introduced 
into each vessel when the same height has been reached. 

384. — Illustrations of Vertical Downward Pres- 
sure. 

If the apparatus (Fig. 220) which is constructed like that of 
Fig. 217, be filled with water, the mercury in the gauges 
will be displaced showing that the pressure due to the weight 



i6i 

of the liquid is transmitted in all directions ; the manometers 
on any horizontal layer will show the same displacement of 
mercury proving that at all parts of the same horizontal layer 
the pressure is the same ; the gauges at different depths show 
increased displacements as the depth increases and if they are 
equally spaced from the hole O in the vertical tube measure- 
ment will show that the length of the mercurial column is 
proportional to the depth or that the pressure is proportional 
to the depth of the surface pressed upon. 

If air is exhausted by means of an air pump from a cylinder 
open at the top the hand held over it will feel the downward 
pressure. 

That the vertical downward pressure on the bottom of a 
vessel is proportional to the specific weight of the liquid may 
be shown by filling two tall cylindrical jars of the same area 
of cross-section, one with water the other with mercury. 

385. — Illustrations of Vertical Upward Pressure. 

The gauge (1) in Fig. 220, shows that the water pressure is 
transmitted vertically upward. 

Another simple piece of apparatus is represented in Fig. 221. 
It consists of a glass cylinder open at both ends and having a 
disk of glass to fit, with a ground joint, at its extremity, the 
disk being held by a string. When the tube is immersed in 
water the disk is held tight to the tube by the upward pressure 
of the water. If the tube is filled with water to the level of 
the surrounding liquid the disk will fall off showing that the 
upward pressure on the disk is equal to the weight of a column 
of water extending from the disk to the level of the liquid. 

The upward pressure of the air may be simply shown by 
placing a card over a tumbler filled with water and inverting 
it. 

386. — Illustrations of Lateral Pressure. 

If a light ball filled with water is suspended from a point 
by a string and the water allowed to escape from a hole in one 
side the ball will be pushed out of its vertical position. 

If a funnel is covered with a piece of sheet rubber and the 



l62 

air exhausted the rubber will be forced in when held in a ver- 
tical position. 

387. — Illustrations of Pressure in all Directions. 

A cask may be bursted by introducing a tube and filling it 
and the tube with water. If air is exhausted from a soap bub- 
ble it will become smaller, retaining its spherical form. The 
bursting of rocks, of water pipes, the crushing of flat bottles 
immersed to great depths in the sea, the water soaking of 
wood at great depths, the limitation offish life, etc., all show 
the fact that water exerts great pressure due to its weight. 

Equilibrium of Liquids. Four cases may be considered. 

388. — 1. A single liquid in a vessel. 

In order that a liquid may be in a condition of equilibrium 

I. Its surface must be everywhere normal to the resultant 
of all the forces acting on the molecules of the liquid. 

For, suppose g (Fig. 222) is a particle of liquid and the sur- 
face is not perpendicular to the resultant. gW, the weight, 
can be resolved into gN and gT. gT on account of the mo- 
bility of the particles of the liquid will produce motion, which 
is contrary to the supposition of rest. Hence, when gT has 
no value the surface will be at rest ; but this can only be 
when it is perpendicular to gW. 

II. Every molecule must be acted on by equal and contrary 
pressiires, otherwise there will be motion in the direction of 
the greater pressure. 

389. — A Level Surface not Necessarily a Plane 
Surface. The surface of the ocean must be convex to fulfil 
the conditions of equilibrium. The depression, in feet, of a 
liquid surface on the earth, below a tangent plane is equal to 
two-thirds of the square of the distance in miles. 

Let BE (Fig. 223) represent the earth, AB its diameter, AT 
a tangent plane and P a point on its surface. PT is the de- 
pression required. Draw PQ perpendicular to AB. Then 

AQ : AP::AP: AB. 

dep. : s : s:: 7912. 

s 2 
dep. in feet = ~ 2 s 2 . 

7912 3 



■63 

The fact that liquids abandoned to themselves assume the 
spherical form and that the surface of water near mountains is 
not truly horizontal also illustrate the above truths. 

390. — 2. A single liquid in communicating vessels. 

Imagine a movable partition whose area is S (Fig-. 224) 
placed anywhere in the vessel. It is obvious that the pres- 
sures on opposite sides of S must be equal, otherwise there will 
be motion. The pressure to the left of S is (§382) F Pl — h^S, 
the pressure to the right is F P2 = h 2 wS, but h^S = h 2 ze/S, 
therefore h, = h 2 , that is, for equilibrium the liquid must stand 
at the same height in both vessels. Practical applications of 
this principle are found in reservoirs for the supply of water 
to towns, in the construction of artesian wells, in the build- 
ing of canals, and the flow of rivers to the sea. 

391. — III. Different liquids in communicating vessels. 

When different liquids are placed in communicating vessels 
the law of balance is that their heights vary inversely as their 
specific weights. 

In Fig. 225 the liquid below HH is obviously in equilibrio. 
The downward pressure on the horizontal plane due to the 
column of liquid on the right is F Pi = h^S^ The downward 
pressure on the left due to the other liquid is F Pz = h 2 w 2 S 2 , but 
for balance these two pressures must be equal, therefore hw 1 S 1 
=h a «/ a S a , but Sj—Sa and therefore h 1 w 1 =h 2 ze' 2 or h 1 : h 2 :: w 2 : zi\. 

It is obvious that the specific weights of different liquids 
may be compared by placing different liquids in the tubes 
allowing balance to take place and comparing the heights of 
the different columns. 

392. — IV. Different liquids in the same vessel. 

If several liquids which do not mix are placed in the same 
vessel balance will be established when each liquid fulfils the 
conditions necessary for a level surface (§388) and they must 
arrange themselves in the order of their specific weights, the 
heaviest being at the bottom. Cream is on top of milk, fresh 
water at the mouth of a river above the salt water and the 
atmosphere rests upon the top of liquids. 



164 

Equilibrium of Bodies in Liquids and Gases.— Principle of 
Archimedes. 

393. — 1. Bodies entirely submerged. 

Every body immersed in a liquid or gas (1) displaces a volume 
of liquid or gas equal to the volume of the body immersed, (2) and 
is subjected to an (a) upward vertical pressure (b) acting through 
the centre of gravity of the displaced liquid or gas, and (c) equal 
to the weight of the liquid or gas displaced. (3). The horizontal pres- 
sures balance each other. §375. This proposition is an immediate 
consequence of Pascal's Principle. 

Imagine the body in the vessel (Fig. 227) to be divided into 
a very great number of vertical prisms. Then any prism is 
pressed down by a force equal to the weight of a column of 
liquid whose base is d e and whose height is a b. It is pressed 
up by a column whose base is fg and whose height is ac. 
Hence there is an excess of upward pressure equal to the 
weight of a column of the liquid whose height is be and whose 
base is fg. But this is the weight of liquid having the same 
bulk as the prism. The same is true of each prism, and 
therefore of the whole body, or calling w the specific weight 
of the liquid, the upward pressure — hwS = fg X ac X w\ the 
downward pressure =hzcS = fg X ab X w. Subtracting the 
second from the first the excess of upward pressure = fg X be 
X w, which is the weight of a bulk of water equal to that of 
the immersed body. 

394. — The truth may be verified by means of the hydrostatic 
balance, (Fig. 228) which is an ordinary balance with one pan- 
suspension shorter than the other. An accurately turned cyl- 
inder of brass is made to fit into a brass cup. The cylinder is 
hung from the bottom of the cup and both are suspended from 
one end of the beam of the balance and equipoised by adding 
weights to the other pan. The cylinder is then immersed in 
water; it loses weight and the amount lost maybe shown by 
filling the cup with water when equilibrium will be re-estab- 
lished; but the volume of the cup is precisely the volume of 
the cylinder, so that the loss of weight is equal to the weight 
of the displaced water. 



i6s 

The same principle may be proved for gases by using 
the baroscope (Fig. 229) which consists of a small weight and 
a large hollow thin sphere having a stop-cock, suspended 
from the opposite ends of a balance. If the whole apparatus 
is placed under the receiver of an air pump and the air ex- 
hausted the stop-cock being closed, it will be seen that the 
large bulb ascends, showing that it was more buoyed up by 
the air than the small counterpoise. If the instrument be 
now removed and the air exhausted from the sphere, the stop- 
cock closed, and the whole apparatus replaced under the re- 
ceiver, when the air is exhausted from the receiver, balance 
will be effected. 

395. — Bodies partly immersed. 

The principle is still true as may be shown experimentally 
by immersing the cylinder (§394) partially and proceeding as 
before. 

It follows that (1) when the w 7 eight of the displaced liquid 
or gas is less than the weight of the body, the body sinks un- 
til (2) the weight of the displaced liquid or gas equals the 
weight of the body, when it remains at rest in any position. 
(3) When the weight of the displaced liquid or gas is greater 
than the weight of the body, it rises and finally floats on top, 
the weight of the liquid or gas displaced by the submerged part 
being then equal to the whole weight of the body. 

396. — These facts may be experimentally verified for water 
by employing the Cartesian diver, and for gases by blowing 
a soap bubble with ordinary illuminating gas and loading 
the bubble with the suds. Balloons are an application of the 
principle. 

397. — It also follows that solids can never float if they are 
heavier, bulk for bulk, than the liquids in which they are 
placed ; that solids which float in one liquid will sink in an- 
other, and that the speed of falling through a liquid depends 
on the size of the body, small pieces falling slowly. 

398. — Floating bodies. 

It follows from the principle of Archimedes that — 



i66 

1. A floati?ig body zvill displace a volume of liquid or gas whose 
weight equals the weight of the body, for a floating body loses 
all its weight. 

2. For equilibrium, the centre of gravity of the body must be in 
the same vertical line tvith the centre of gravity of the displaced 
liquid or gas {called the "centre of buoyancy"); for, the weight of 
the body acting through its centre of gravity presses it down, 
and the resultant upward pressure through the "centre of 
buoyancy" presses it up ; these for balance must be equal and 
opposite. §21. 

3. For stable equilibrium the metacentre must be above the cen- 
tre of gravity of the body ; for tinstable equilibrium below and for 
neutral at the centre of gravity. 

Let A (Fig. 230) be a vessel of water in which a floating 
body B is placed, let g be the centre of gravity of the body, 
and F 2 its weight. Represent the resultant upward pressure 
by Fj. In Fig. 1 F, and F 2 must be equal and opposite for 
balance. §21. Now tilt the body so that it occupies the po- 
sition in Fig. 2 ; the centre of gravity, g, of the body remains 
unchanged, but the centre of gravity of the displaced water 
has changed to the position C. The point M, where the line 
of pressure through the new centre of gravity of the displaced 
water intersects the axis of the body (determined in Fig. 1), 
is called the " metacentre." In Fig. 2 it is obvious that F, 
and F 2 will tend to right the body, in Fig. 3 to tilt it still 
more. §27. 

399.— The tendency to right the body in Fig. 2 is measured 
by the moment of the couple F^. Hence, if the angle of dis- 
placement gMC is 0, FjL, = Fj • CM sin. 0, which is called the 
moment of stability of the floating body. This equation shows 
that this moment (1) increases with the weight of the body, (2) 
with the distance of the centre of gravity below the metacen- 
tre, (3) with the angle of inclination. 

Hence the force with which the body recovers its position, 
called the moment of stability, is increased, (1) by using ballast, 
(2) is diminished by passengers rising, (3) and if g is not ex- 



167 

actly over the keel the vessel will tilt, as is shown when the 
cargo shifts. It follows that life preservers should be placed 
around the neck. 

400. — Specific Gravity. Definition. The specific gravity of 
a body is the weight of one cubic foot of it compared with the 
weight of one cubic foot of some other substance taken as a 
standard, or it is the weight of a body divided by the weight 
of an equal volume of the standard. These two definitions 
are evidently the same since the weight of a body, W n is equal 
to its volume, V n multiplied by its specifie weight w xi or W x 
= Vw^ and the weight of an equal volume of the standard 

equals W, = V,w Q , and therefore -— = ~ — - — — . 

W 2 V Q w, w 2 

The standard for solids and liquids is pure water at the tem- 
perature of its maximum density, 4°C or. 39. i°F. Air or hy- 
drogen at o°C and under barometric pressure of 760 mm. is 
taken as the standard for gases. 

The ratio of the densities of bodies is the same as the ratio 
of their specific gravities. For (§9) if m, is the mass of a body 
whose volume is V^ and density 6 iy then n^ = V^, ; if m 3 is 
the mass of a standard volume V 2 whose density is 6. 2 , then 

m 9 = V^ 2 ; therefore — L -- — *, that is, the ratio of the densities is 
6 2 in, 

the same as the ratio of the masses. But (§266) m, — — ■ = V, — ■ 

■■0.9 

and m„ = — ? = V„ — -, therefore, — - = — -, that is, theratio of the 
9 'ff '«>. m.' 

/? iv 
specific weights is the same as the ratio of their masses] hence 1 __ — l 

that is, Hie ratio of the densities is the same as the ratio of the specific 
iveights or specific gravities, a fact which was proved experi- 
mentally by Newton. 

If ft, or the mass of unit volume of pure water at 4°C is taken 
as unit density — ' ==^ 1} and a table of specific gravities is the 
same as a table of densities. 



i68 

Methods of Obtaining Specific Gravities of Solids. 

401. — I. Of solids heavier than water. 1. The hydrostatic balance 

(§394) may be employed to determine the specific gravity of a 

body heavier than water. Since specific gravity is the 

weight of a body 

the n under- 



weight of an equal volume of pure water at 4°C' 
ator may be found by determining the weight in air directly 
with the balance ; the denominator by substituting for it the 
loss of weight the body sustains when submerged in water ; 
for by the principle of Archimedes (§393) this loss of weight is 
equal to the weight of an equal bulk of water. 

2. Nicholson's hydrometer is represented in Fig. 231. D 5 and 
D 2 are pans attached by stems to a float B loaded so that the 
whole apparatus stands vertically when placed in water. 1 is 
a file mark on the upper stem so placed that when the instru- 
ment is loaded with 1000 grains it will be at the level of the 
pure water in which the instrument is immersed. Put the 
solid, the specific gravity of which is to be determined, in the 
upper pan and add standard weights until the instrument 
sinks to the mark 1 ; then 1000 minus the weights added will 
give the weight of the body in air ; remove the body and 
place it in the lower pan ; add weights until the instrument 
sinks to 1. The weights added will give the weight of an 
equal bulk of water. If the lower pan is removed it is called 
Fahrenheit's hydrometer. 

402. — II. Of solids lighter than water. 1. If the hydrostatic bal- 
ance is used the light body may be attached to a heavy one 
which will sink it in water and whose weight is known. The 
loss of weight of the combination less the loss of weight of the 
heavy body alone will give the weight of a volume of water 
equal to the volume of the light body. Divide the weight of 
the body in air by the latter quantity and the result will be 
the specific gravity. 

2. When Nicholson's hydrometer is employed a cage is sup- 
plied with the instrument so that the light body may be kept 
submerged. 



169 

403. — III. Of solids soluble in water. With the hydrostatic bal- 
ance find the specific gravity in some liquid in which the body 
is not soluble ; also the specific gravity of this liquid relatively 
to water. §405. The product of the two specific gravities 

will be the specific gravity required. Let — be the specific 

w 
gravity of potassium in naphtha and — - the specific gravity of 

w w 
naphtha relatively to water, then — - 1 X — -will be the specific 

gravity of potassium with water as the standard. 

404. — IV. Bodies in the form of powder. With the hydrostatic 
balance weigh the powder in air. Place a flask (§403, 2) filled 
with water, and the powder on one pan and add weights until 
balance is established. Put the powder in the flask and wipe 
off the displaced water. The weight necessary to establish 
balance again is the weight of an equal bulk of water. Divide 
the weight in air by the latter quantity and the result is the 
specific gravity required. 

Methods of Obtaining the Specific Gravity of 
Liquids. 

405. — 1. The hydrostatic balance may be employed by sus- 
pending from the bottom of the short pan, by means of a fine 
thread or wire, a ball of glass or other material which cannot 
be chemically acted upon by the liquid. Immerse the ball 
first in the liquid the specific gravity of which is to be deter- 
mined. Note its loss of weight. Then immerse the ball in 
pure water and note the loss of weight. The former quantity 
divided by the latter will give the required specific gravity. 

2. The specific gravity flask is made of glass and fitted with a 
stopper having a capillary hole bored through it. The bottle 
will hold 1000 grains of pure water. An equipoise to balance 
the weight of the bottle accompanies each instrument. Fill 
the bottle with the liquid, weigh it, divide its weight by 
1000 and the result is the specific gravity since the quotient 
is the ratio of the weights of equal bulks. 



170 

3- — Nicholson'' s hydrometer (§399, 2) may be used to find the 
specific gravity of a liquid if the weight W of the instrument 
is first determined. Place the hydrometer in water and add 
1000 grains ; it will sink to the notch on the stem. Place it 
in the liquid, the specific gravity of which is to be determined, 
and add weights w until the file mark is at the level of the 
liquid. The same volume has been displaced in each case. 
-p. W+w _ weight of the liquid displaced 

W -4- 1000 weight of an equal bulk of water 

4. The method of difference of level has been described in 
§39 T - 

5. Specific gravity bulbs are small glass balls which may be 
thrown into a liquid. The one which remains suspended in- 
dicates by the number on it the specific gravity. 

406. — Methods of Finding the Specific Gravity of 
Gases are given, with the proper precautions, in works on 
heat. 

407. — Tables of specific gravities are useful to distinguish one 
substance from another having a similar appearance. The 
weight of a body may be calculated from the volume when its 
specific gravity is given. 

408. — Scale Hydrometers. In Nicholson's hydrometer 
(§399) the volume of the liquid displaced is a constant quan- 
tity and the weight of the instrument varies. In all such cases 
the weight varies as the specific gravity ; for, let V, be the 
volume displaced when the hydrometer is placed in a liquid 
whose specific weight is w x , then W^V,^, ; also, let V 2 be the 
volume displaced under like conditions for a liquid whose spe- 
cific weight is ze> 2 , then W 2 =V 3 ze>„, but V, = V, . " . W, : W 2 :: 

Wj : ze> , or — - = — -— sp. gr. But hydrometers may be so 

W 2 w 2 ' 

constructed that while the weight of the instrument remains 
constant it sinks to different depths in different liquids, or the 
volume varies. Such instruments are called scale hydrometers. 
The principle of Archimedes applies to them. For when a 
floating body is placed in a liquid it sinks until the submerged 



I 7 I 

part displaces a volume of liquid whose weight equals the 
weight of the body. §393. Hence with a given floating body 
if the liquid is heavy, the descent will not be so great as in a 
light liquid. If V, represents the volume of liquid displaced 
whose specific weight is a;,, then W l =V 1 w l . IfV 2 is the volume 
of another liquid ze/ 2 displaced by the same instrument W 2 =V 2 ze/ 2 , 
but W, = W 2 and V, : V 2 :: w 2 : w 19 that is, the specific weights 
vary inversely as the volumes displaced. If one of the liquids is 

pure water 6 l = l and 6 2 = —~. 

1. Gay Lussacs volumeter (Fig. 232) consists of a cylindrical 
float graduated in equal divisions. When placed in pure water 
the level is at 100. If it were placed in another liquid and 

100 , r 

stood at 60, then 100 : 6o::sp. gr. : 1, or sp. gr. = --=1.00. 

2. Baumes hydrometer is graduated arbitrarily. For liquids 
heavier than water, the instrument is floated in pure water and 
the level near the top marked o°. (Fig. 233. ) To obtain other 
scale divisions the instrument is floated in a mixture of salt 
and water, 15 of the former to 85 of the latter. The level is 
called 1 5 . The intermediate distance is divided into 15 parts 
and the graduations are continued to the bottom of the in- 
strument. When used for liquids lighter than water it is first 
floated in pure water and the level near the bottom marked 
io°. (Fig. 234). The zero of the scale is determined by 
floating it in a solution of salt and water 10 to 90. The di- 
visions are then numbered up from zero. 

Gay Lussac's Alcoometer (Fig. 235) differs from Baume's 
simply in the graduation. The mark corresponding to the 
level of pure water is o°. When placed in absolute alcohol 
the level is marked ioo°. The interval between o and 100 is 
graduated by placing the instrument in mixtures of alcohol 
and water, containing 10, 20, 30, 40, etc., percent, of alcohol. 

Various other instruments, called salimeters, lactometers, 
vinometers, urinometers, are constructed on the same plan. 

409. — Pressure of Gases. The formula R p = hz£/S, (§382) 
would apply to pressures exerted by the atmosphere upon a 



172 

surface, but it is not possible practically to obtain a value for 
h. The air being compressible the lower layers are more 
dense than the upper. When in equilibrium the pressure on 
each horizontal layer is the same and the weight of the air is 
transmitted in all directions. §384. 

It is obvious that if the pressure on each square inch of sur- 
face were known the total pressure R p would equal the pres- 
sure on each square inch multiplied by the number of square 
inches in the surface or R p = pS. This is true provided the 
pressure on each square inch is constant. If it were varying 
the surface may be imagined to be divided into parts so small 
that the intensity over each may be regarded as constant. The 
total pressure or R p will then become -pdS. 

An experiment performed by Torricelli in 1643 gave the 
pressure exerted by the air on each square inch of surface. 
He took a glass tube (Fig. 236) about 36 inches long, closed at 
one end and filled it with mercury. Holding the finger over 
the open end he inverted the tube in a cistern containing mer- 
cury. The mercury oscillated in the tube and finally came to 
rest at a height about 30 inches above the surface of the mercury 
inthecistern. The column was sustained by the air pressure on 
the mercury in the cistern. §387. If the tube has a sectional 
area of one square inch the weight of the column is about 15 
pounds, which represents the pressure of the air upon each 
square inch of surface. Since the specific gravity of mercury 
at 32°6 F is 13.59, tne a * r would sustain a column of water 
33.91 feet high or 10.336 m. This pressure, 15 pounds to 
the square inch is technically called " an atmosphere." 

The barometer is this simple apparatus of Torricelli's, fur- 
nished with a scale and is used to determine simply the varia- 
tion in the pressure of the atmosphere. The average pres- 
sure at the sea level is 14.7 pounds per square inch, which cor- 
responds to a height of a column of mercury of 29.922 inches 
or 760 mm. 

410. — Compressibility of Gases. 

Since gases tend to expand (§374) indefinitely they are said 
to exert a tension ; the force which they will sustain is also 



173 

called the pressure of the gas. The relation between prrs- 
sure, density, temperature and volume may be expressed by 
the laws of Boyle and Charles. 

411. — Boyle's Law. If the temperature is kept constant the 

1 

volume of 'a gas varies inversely as the pressure orV 00 — -; since 

with a constant mass the volume is inversely proportional to 
the density (§9), The law may also be stated as follows : 

If the temperature is kept constant the density varies direct- 
ly as the pressure 6 go F p . The law may be experimentally 
proved by using a bent tube (Fig. 237) having a long and a 
short branch, the latter being closed. Mercury is poured into 
the long branch until it stands at the same level HH, and just 
closes the bend. The tension of the enclosed air is now that 
of the atmosphere. If more mercury be poured in the air in 
the small branch will be compressed and its tension will be 
measured by the pressure it will sustain. It is found that 
when the volume has been reduced to one-half its original 
value, the pressure is that due to a column of mercury 30 inches 
high plus the pressure of the air on top of the mercury or two 
atmospheres. When the volume is reduced to one-third the 
pressure is that due to 60 inches of mercury plus the atmo- 
spheric pressure or three atmospheres. Hence double pres- 
sure corresponds to half volume, treble pressure to third 
volume, which is the law. 

For pressures less than atmospheric a straight tube (Fig. 
238) a yard in length, closed at one end, may be immersed in 
a deep cistern containing mercury. The tube is almost filled 
with mercury, the finger is placed over the end and the tube 
is inverted in the cistern. If the tube be now placed so that 
the level of the mercury is the same within and without, it is 
obvious that the tension of the enclosed air equals the exter- 
nal air pressure. If the tube be drawn up so that the volume 
of the imprisoned air is doubled it will be found that in the 
tube there is a column of mercury fifteen inches high. Now 
the tension of the air plus the height of this column is balanced 
by the atmospheric pressure, hence the tension of the air is 



i74 

one atmosphere minus half an atmosphere, or by doubling 
the volume the pressure has become one-half an atmosphere. 

412. — The Law of Charles, sometimes called GayLussac's 
Law. 

If a given volume of gas is enclosed in a vessel of constant 
size and the temperature raised the tension is increased ; if 
enclosed in a vessel having a movable piston, that is under 
constant pressure, an increase of temperature produces in- 
creased tension. The exact law may be stated as follows : 

If the density is kept constant the tension increases with the in- 
crease of temperature ; if t lie pressure is kept constant the volume 
increases with the increase of temperature. 

The law may be mathematically expressed as follows : Let 
V to be the volume of a gas at the temperature of melting ice 
0°C, or 32°F, and V tl be the volume of the same gas at the 
temperature t„ let x be the increase in volume of unit volume 
for one degree rise of temperature, called the "co-efficient of 
expansion" of the gas, then 

V h = V t0 + V to *t, = V t0 ( 1 + «t, ) = V„ ( 1 -t- .oo 3 66t, ) - 
V to [I+.C0204 (tj — 32 )]. From this equation the volume at 
o° may be calculated if the volume at t x ° is given. 

Again let V ta be the volume at t 2 ° then 

^ V t! =V to +V,„*t ! =V to (i+*0=V, (l+.oo 3 66tJ=V !o [l +.00204 

(t 2 — 32 °)~\. This equation enables us to calculate the volume 

at o° if the volume at t,, is given. 
V^ _ Vt (l + *tj = 1 -f xt 1 _ l + .OQ 366t 1 _ 1 +.00204(1 -— 32 ) 
V t , ~~ V to (i + xt j 1 + yX~ 1+ .003661," 1+. 00204(^-32°) 
This equation makes it possible to calculate the volume at 

t a if the volume at t, is given. 

Since by Boyle's Law (§411) the volume varies inversely as 

the density, at constant temperature, 

Vt t = 6, = i + . 0366^ = 1 + . 00204(^- 32°) 
Vt 2 6] l+.oo"366jt 3 i+.oo204(t 2 — 32°) 
This equation renders it possible to calculate the density at 
t 8 when the density at t x is given. 



i75 
Again if the tension changes 

V tl =(l-f.oo 3 66t 1 )P^V o 

Ptx 

V f2 -(l+.oo 3 66t 2 )^_V o 

Pt 2 
Vt, _ l + .0 0366^ pt L 

V t /~ i-f,oo 3 66t 2 " p tl 

If the co-efficient of expansion of a gas is = 00166 it 

273 * ' 

follows that if the temperature is reduced i° a cubic foot of 

air would become 1 ; if 2 , 1 — 2 ; if ioo° 1 — -— ? 

2 73 273 273 

andif273°, \—^2A r zero. This point is characterized 

2 73 
by the absence of gaseous elasticity or heat and is there- 
fore called the absolute zero. All temperatures measured from 
absolute zero are called absolute temperatures and may be 
represented by t. 

• The relations between volume, pressure and temperature 
can be more simply expressed by using this idea and the laws 
become 

1. If the pressure is kept constant the volume varies as the 
absolute temperature. 

2. If the volume is kept constant the pressure varies as the 
absolute temperature. 

3. The product of the pressure and volume divided by the 
absolute temperature, for all changes, equals a constant, or 

t 

Boyle's Law follows from (3) when t is constant or p °°yt 

Charles's Law follows from (3); when Vis constant p 00 t, when 
p is constant V cot. 

413. — Kinetics of Liquids and Gases. The Principle of 
Continuity. A liquid or gas is supposed to flow on continu- 
ously, that is, without perceptible breaks or change of density. 
If this is true 



\j6 

The amount of the liquid or gas which passes a cross section 
in a second anywhere in the stream, technically called the flow 
per second or strength of current is the same. 

Leonardo da Vine? s statement of the law. It is obvious, 
then, that the discharge through any cross-section in one sec- 
ond is a cylinder whose area of cross-section is that of the 
stream at that point and whose length is the distance traversed 
by the stream in one second or the velocity. s=Tv=l X v=v. 
L,et d be the flow per second through an area a, and v, 
the velocity ; also let C 2 be the flow per second through 
another area a 2 and v 2 the velocity ; then C 2 — a^, ; C 2 — a 2 v 2 ; 
but C, = C 2 ; therefore a a v 2 = a s v 2 or v, : v 2 :: a 2 : o. 1 that is, the 
velocity of flow varies inversely as the area of cross-section of 
the stream. 

414. — If friction between the molecules themselves of the 
liquid or gas, and between the molecules and the walls of the 
canal were absent the stream would flow on forever, (§258, 1) 
but these resistances do exist, therefore force has to be exerted 
to keep up the flow. §261. 

A column of liquid or gas at rest may be set in motion by 

increasing or diminishing the pressure at any point. Let the 

pressure at A, be Pa„ and at another point A 2 be Pa 2 . The 

difference of pressure or difference of level between A 1 and 

A 2 will be Paj — Pa 2 , and the difference of pressure per unit 

p p 

length will be — * ^» The force which produces the flow 

A,A, 

depends on this expression. Now F= mv (§261) hence F= 

P A * - Pa 2 
mv = — — - — -. 

AjA 2 

A stream may be imagined to be filled with lines which rep- 
resent the direction of motion of the molecules at every point. 
§419. Such lines are called stream lines or lines of flow. §346. 
In a steady stream the lines at any point are supposed to be 
parallel and equally spaced. If the stream widens the lines 
of flow diverge ; if it narrows they converge. The facts 
already given in reference to lines of force apply to lines of flow. 



i 7 7 

Tf the unit mass is moved the measure of the force is the 

velocity, which depends on the difference of pressure per unit 

j." ta Paj — Pa 2 

distance. F = v = — - • 

Cor. If the difference of pressure is small the velocity will 
be small for a given distance between A 1 and A 2 . 

If the velocity is great and the difference of pressure is unity 
the distance between A 1 and A 2 must be small. Hence, the 
theory of flow resembles the theory of potential. §345. 

415. — The volume of liquid or gas discharged from an orifice 
in a vessel is, obviously, the discharge per second multiplied 
by the time in seconds, or, if Q is the volume required and C 
the efflux per second Q = CT. Now C can be obtained (§413) 
by multiplying the area of the cross-section of the stream by 
the velocity or C — av, therefore Q = avT, that is, for a steady 
stream the volume discharged is equal to the product of the area of 
cross-section of the stream, the velocity of flow and the time in 
seconds. 

416. — Head of Water. When a liquid issues from an aper- 
ture in a vessel the force which produces the motion is the 
hydrostatic pressure on the aperture, which is equal to hz^S 
(§382) and which is proportional to the depth of the orifice 
below the free surface of the liquid or the " head." 

417. — Torricelli's Law. Under constant pressure the velocity 
of flow from an aperture, small compared with the vessel, is the 
same as a body would acquire in falling from the free surface to 
the aperture. 

This is an immediate consequence of the principle of the 
conservation of energy. Let a vessel (Fig. 239) be kept full 
of water to the height HH so as to produce a constant velocity 
of efflux. An orifice O is made in the bottom from which 
water escapes. Let a 1 be the area of cross-section of the orifice 
and a. 2 the cross-section of the vessel at HH. Let v, be the 
velocity of efflux at O and v 2 the velocity of sinking at HH. 
Let m be the mass of the liquid which issues from O in one 
second under the head h or the vertical distance of O from HH. 



i 7 8 

The kinetic energy of m as it flows from O is ^mv', and as 
it sinks below HH, ^mv; 2 . Hence there has been a gain of 
kinetic energy equal to l / 2 m (V— v 2 ) ; but the work performed 
by m to produce this increase is m#h. But these two expres- 
sions are equal (§343) or 

y 2 m (v"— v^) = m#h. 

Now by da Vinci" s statement (§413) v, : v 2 :: a s : a } or v 2 = 

a. v 

-i— J , which substituted in the last equation gives 
a 2 

~2ffh~ 



1 — - 1 - 

a 2 2 

a 2 
If a, is very small compared with a 2 -^ will be small com- 

a 2 

a 2 
pared with one and 1 — -—will ultimately become one, hence 



v, = T/2^7h, but this is (§204) the velocity a free falling body 
would acquire in falling through the distance h or the head. 

Cor. Since the velocity of efflux varies as the square root of 
the head and the head varies as the hydrostatic pressure, the 
velocity also varies as the square root of the pressure. 

418. — Since the discharge varies as the velocity (§415) and 
the velocity varies as the square root of the head the discharge 
varies as the square root of the head. This fact may be illus- 
trated by the u spouting apparatus" represented in Fig. 240. 
It consists of a tall vase having holes in the sides at equal dis- 
tances from each other. The vessel is kept full of water by 
arranging an overflow at the top. Various shaped nozzles and 
pipes may be attached to the lateral orifices. If a quart mea- 
sure be held under nozzle number 4, and a pint one under 
number 1 and the water allowed to escape at the same instant 
the two vessels will be filled in the same time. Hence, the 
quantity discharged is proportional to the square root of the 
depth of the jet below the surface, since the nozzle number 4 
is four times the distance of number 1 from the surface. 



179 

419- — Discordance between Theory and Practice. These 
equations will not give the actual discharge. If a quantity of 
bran or other substance having almost the specific gravity of 
water be introduced into a vessel from which a liquid is es- 
caping through an orifice the particles will arrange themselves 
in "streamlines" and some of the causes of the discordance 
between theory and practice may be observed. 

i. The Contractu Vein. It will be seen (Fig. 241, 1) that 
the jet is not a perfect cylinder but conical in shape and con- 
stricted at a short distance from the orifice. This constricted 
portion is called the "vena contracta." Its conical form is 
accounted for by the convergence of the liquid filaments towards 
the aperture. Beyond the contracted vein the jet is nearly cyl- 
indrical. In making calculations, therefore, on the efflux the 
area of the aperture must not be taken but the actual area must 
be multiplied by some quantity called the "coefficient of contrac- 
tion" which will give the effective area ; it may be represented 
by K. Taking this fact into consideration the discharge per 
second must be C = (A'a)v in which a is the area of the cross- 
section of the orifice, instead of C==av. 

420. — 2. The velocity of actual discharge is not the same 
as the theoretical, hence, instead of using v as the actual ve- 
locity it must be multiplied by some quantity which will make 
it equal to the real velocity. Such quantity is represented by 
$ and is called the " coefficient of velocity." Hence, introdu- 
cing the two corrections, the discharge per second becomes 
C = (Aa) (0v) instead of C = av. 

421. — 3. Experiment also shows that the actual flow C is 
not the same as the theoretical ; hence C must also be multi- 
plied by a quantity which will produce the real flow. Such 
quantity is represented by fi and is called the " coefficient of 
efflux." Hence, introducing the three corrections it is found 
that the real discharge per second is fiQ, == (Aa) (#v) = A"$ (av) 

-= /(0a [/ 2Qh 5 but C — av, hence /x — A'^ 5 or the coefficient of 
efflux equals the product of the coefficients of velocity and contrac- 
tion. Hence, the actual discharge per second may be repre- 
sented by / «a l/ / 2 ( ^h. 



i8o 

422. — Mouthpieces. The amount of outflow from an orifice 
depends very much upon the form of the mouthpiece as well 
as on the position of the orifice. Fig. 241, 2. If the ajutage is 
made in the form of the jet and ending at the vena con- 
tracta (Fig. 241, 3) the discharge will equal avT, because the 
effective area is the area of the vena contractu. If the mouth- 
piece projects inwards both Q and v will be diminished. Fig. 
261, 4. If a cylindrical mouthpiece projects outwards and the 
tube is filled the discharge will be greater than when the orifice 
is without a mouthpiece, provided the tube is wetted by the 
liquid. 

423. — Efflux of Liquids from Orifices in the Side of a Ves- 
sel. If the gates in the side of the spouting apparatus (Fig. 
242) be opened each particle of the liquid, being urged forward 
by a constant force and at the same time acted upon by gravity, 
will describe a parabola. §210, 2. 

Since each particle of the liquid is projected horizontally, the 

x 2 

equation of the path of the projectile y = x tan. 6 ■ - 

H F r J J 2V 1 2 COS. 2 tf 

§21 1 becomes y= — - — , since = o, and therefore tan. 6 = 
2Vj 2 

and cos. = 1. Hence making v, 2 -- 2$h and calling the 
downward ordinates plus x 2 = 4hy. 

In Fig. 242 if the jet is taken at the nozzle J, , h is LJ X . 
If any other horizontal plane ox is taken then the equation 
becomes x 2 = 4hy = 4b ■ J,o = 4hh/. 

These two equations give the motion of the jets. 

424. — If a semi-circle be described upon the side of the vessel 
as a diameter and ordinates be drawn from each outlet to the cir- 
cumference, the horizontal range on the level of the base of a 
jet will equal twice the ordinate drawn from the outlet to the 
circumference. 

For, taking J l5 from the property of the circle J, H, = 
VXO • LJ : - i/h'h orhh' = .LH7- 

..•;x— 4hh' = 4j 1 H/ 

x -- 2JJHJ or since x = ox 5 

OX 5 = 2J.H, 



i8i 

425. — Cor. Hence it follows that the horizontal range of the 
middle jet is greatest, and of two jets equally above and below 
the middle one the horizontal ranges are equal. §211, 4, 5. 

426. — Flow in Open Channels. Observation shows that 
water in a stream, canal or other open channel is steady, that 
is, the quantity which flows past a given cross-section in the 
same time is constant in all parts of the stream. The mean 
velocity therefore is equal to the discharge per second divided 
by the area of cross-section. The friction at the bottom and 
sides of the channel is greater than that between the mole- 
cules of the water itself. It follows that the velocity varies 
at different parts of any cross-section and that there is a line 
at the upper surface near the middle of the stream which has 
the greatest velocity. The quantity discharged in a given 
time may be determined experimentally by finding the width 
and average depth by taking soundings ; their product will 
give the area of cross-section; by finding the average velocity 
by taking the mean velocity of various light bodies floating 
on top and at various depths below the surface and by substi- 
tuting the values thus determined in the equation Q == avT. 

427. — Constant Flow of Liquid in Uniform Rigid Pipes. 

The flow of water through pipes is determined by certain em- 
pirical formulae which can be found in practical treatises on 
hydraulics. The veriest elements of the subject are here 
given. Let a vessel (Fig. 243) of height H be kept filled with 
water. At the base a uniform rigid pipe P,P 2 is attached. 
At various points along this pipe vertical glass tubes are 
inserted. The water is allowed to escape from the aperture P 2 
of the pipe. The water will issue from the aperture with veloc- 
ity v. Call the height due to the velocity v, h v , or velocity- 
2 

head. Then h v = — . §417. It is also observed that the 

water stands at various heights in the piezometers. The 
height at any point indicates the amount of resistance which 
is yet to be overcome as the water seeks the outlet. It is ob- 
served that, while h v is constant throughout the whole length 
of the pipe, the head which overcomes the resistance varies 



182 

from point to point. Call this head h p or the pressure-head. 
If there were no other resistances the total head 
H = h v + ,h p (i) 

K = ?- (2) 

Haagen has shown that the resistance to flow in a pipe may 
be expressed in equivalent head of water by the equation 

h p =R — 1 1 Cj — +c 2 —J (3) in which 1 is the length of the 

pipe, d is its diameter, v the velocity of flow and c l and c 2 
constants to be determined by experiment. In any pipe of 1, 
d dimensions and constant head H the velocity will be such as 
to require that these three equations shall be true. 



i«3 

Apparatus Founded Upon the Preceding Principles. 

428. — The Barometer, §409, requires special precautions in 
its construction. The tube must be rilled with pure mercury 
and every trace of moisture and air removed. After the tube 
is filled certain corrections must be made in order to obtain 
the height due to the pressure of the air. 

429. — These corrections are : 

(1) for Capacity. In the ordinary barometer with a rigid 
cistern it is evident that when the column falls in the tube the 
level rises in the cistern and vice versa. Now the height is 
to be measured from the level in the cistern to the top of the 
column, but the cistern level is changeable. A correction 
must therefore be calculated 

(2) for Capillarity. When a glass tube is filled with mer- 
cury, which does not wet it, the mercury is depressed by capil- 
lary attraction. Hence, to get the true height due to the at- 
mospheric pressure alone, a correction must be added to com- 
pensate for this depression. Tables have been constructed 
showing the corrections for different diameters of tube and for 
different heights of the convex portion of the column. When 
the column rises the convex portion or meniscus is higher, 
when falling, lower, hence the column should always be read 
on falling before applying the tabular corrections. 

(3) For Temperature. A warm column of mercury exerts 
less pressure than a cool one of the same height, since heat 
produces expansion. It is customary to reduce all readings to 
32°F or o°C. If the reading is above 32 the correction is 
substractive, if below, additive. As the metal scale is also 
affected by heat a correction must be made for its variation in 
length. 

(4) For Index Errors. Index errors are those made in the 
graduation of the scale and in the position of its zero. These 
errors are determined by comparison with a standard barome- 
ter and a table is constructed for each half inch of the scale. 
They are constant. 

(5) For Difference of Altitude. Since the barometric col- 



184 

umn is sustained by the pressure of the air and the amount of 
air decreases as the barometer is carried to greater altitudes, 
to make all the readings consistant it is necessary to reduce 
them to what they would be at some level agreed upon. The 
sea level has been selected for this purpose. 

(6) For Unequal Intensities of Gravity. It is obvious from 
§269 that if two stations are at different distances from the 
earth's centre the attraction of gravitation will differ. Hence, 
the pressure of the air, which depends upon the weight of the 
column will differ, but the weight by §266 depends upon Q or 
" the acceleration due to gravity." To make the readings of 
the barometer comparable a correction ought therefore to be 
made for Q. 

430. — Fortin's Barometer is constructed to avoid the cor- 
rection for capacity. If the tube has a wide bore the correc- 
tion for capillarity is avoided. It consists (Fig. 244) of the 
usual barometer tube drawn to a point and immersed in a cis- 
tern of glass so that its level may be seen. Attached to the 
lower part of the cistern is a flexible bag. A rigid case en- 
closes the bag and a screw passes through the bottom of the 
case. An ivory or steel pointer is screwed into the top of the 
cistern. To read the instrument the screw is raised or lowered 
until the surface of the mercury just touches the point of the 
pin. A vernier is used in connection with the scale to 
measure small divisions. This arrangement enables a reading 
to be made when the mercury is at the same lower level. To 
transport the instrument the mercury is screwed up and the 
instrument inverted. 

431. — The Siphon Barometer while convenient has its dis- 
advantages. Two observations have to betaken and the mer- 
cury becomes impure from contact with the air. There is no 
error of capacity. It is a barometer tube (Fig. 245) bent upon 
itself with the lower end open. It is obvious that the height 
of the column required is the distance from the lower level to 
the upper level of the mercury. 

It is obvious that barometers should be held in a vertical 



1 8 5 

position when readings are taken. Hence it is sometimes sus- 
pended on gimbals like a mariner's compass. 

432, — The Aneroid Barometer. A metallic barometer has 
been constructed for convenience of transportation. It is a 
cylindrical box (Fig. 246) with a corrugated top from which 
the air has been almost completely exhausted. A spring pre- 
vents the crushing of the box by atmospheric pressure. As 
the pressure varies the top of the box is pushed in or 
thrown out. These motions are communicated by suitable 
mechanism to an index which moves over a graduated scale. 
The instrument is set by comparison with a mercurial stand- 
ard barometer. The errors in this instrument are apt not to 
be constant on account of molecular changes in the spring — 
hence frequent comparisons with the standard are necessary. 

433. — The Air Pump was invented by Von Guerike in 1650. 
It is used to exhaust air from closed vessels. In its simplest 
form — the single barrel pump — it consists (Fig. 247) of a metal 
cylinder, C, having a hole in the bottom in which is a valve, V, 
opening upwards. In the cylinder is fitted an air-tight piston, 
P, free to move along its length. The piston also contains a 
valve, V 2 , opening upwards. A connecting pipe, T, extends 
from the valve V 1 to the centre of a circular plate, M, upon 
which the vessel, R, to be exhausted is placed. The joint 
between the glass receiver and plate is made air-tight by 
grinding. A stop-cock, S, is placed in the pipe, T, to cut off 
connection between the receiver and the cylinder when de- 
sired. An additional valve is also placed in the connecting 
pipe to admit air into the receiver when it becomes necessary 
to remove it from the plate after exhaustion has been effected. 
The piston may be operated by the hand, a lever, a crank or 
some other mechanical contrivance. 

434. — Mode of Operation. 

Imagine the piston to be at the bottom of the cylinder. 
Both valves are now closed by their own weight. If the pis- 
ton is drawn up, its valve, V 2 , will be kept closed by the at- 
mospheric pressure on top (§384). The portions of the ma- 



1 86 

chine occupied by the air, namely, the receiver and connect- 
ing pipe have become enlarged by the volume of the cylinder 
below the piston on account of the upward motion of the pis- 
ton and the air on account of its indefinite expansibility (§374) 
will expand, open the lower valve, V n and fill the enlarged 
space. The enclosed air has thus increased in volume by an 
amount represented by the volume of the barrel and its density 
has been correspondingly diminished. If the piston is now 
pushed down the valve, V„ will close on account of the air 
pressure on it and the air having no other way to escape will 
open V., and pass into the external air. One barrelful of air 
has thus been removed from the receiver. If the piston is 
again moved up and down the same operation will be repeated. 

435. — A Perfect Vacuum Impossible. 

It is theoretically impossible to obtain a perfect vacuum by 
means of the air pump. An application of Boyle's Law (§411) 
will show this fact. Let V r be the volume of the receiver and 
connecting pipe of an air pump and V c the volume of the cyl- 
inder. Let 6 1 be the density of the enclosed air at the begin- 
ning of the experiment, and 6 2) 6 9i &c. the densities after the 
first, second, third, &c. strokes. Then since the air which 
originally occupied the volume V r after the first stroke of the 
piston occupies V r + V c , 

V r + V c : V,::^:^ 

(V r + V c )£ 2 = V r .^ 

Hence 6 2 = 6, . - J^L_ - 
V r + Vc 

Similarly 6 3 = ? K . __ Vr = 6 ' V 



v r + V c ' W r -f v 2 

\Vr + Vc) \ V r + Vj 

v r 



D n = 0, 

1 \V, + Vc/ 
If these values be arranged in a line as follows : 



1 8 7 
V r \ g I V r 



J'^lv^vJ 3 MvJvJ 



^v r + v c /' w r + v 

it is seen that they form a geometrical progression ; from 
which it is evideut that 6 n will never become zero unless n is 
infinite, which is impossible. 

Practical illustration. Suppose the volume of the cylinder 
is one-tenth that of the receiver and the connecting pipe, and 
that 6 l is the original density of the air, the parts remaining 
in the receiver after each stroke will be 

6 6 ■ 9 6 . 8l g .729_ g . 6561 6 . 59Q49 &c 
IO IOO IOOO IOOOO 100.000 

If the cylinder is very large compared with the receiver and 
connecting pipe a few strokes will practically complete the 

exhaustion. Let V c = ioV r , then the series will be 8 iy 6 X 



11 



6 1 • JL , 6 r -J—, 6 t ■ - , &c. 

121 1331 H774 1 

Practically, there are so many mechanical difficulties in the 
construction of the air pump that the actual vacuum obtained 
is very different from the theoretical one, for a given number 
of strokes. 

436. — Defects of the Air-Pump. 

I. The elastic force of the air is not sufficiently strong, after 
a few strokes, to open the lower valve. To overcome this diffi- 
culty the valve is made very light. Oiled silk is usually em- 
ployed. 

It is also obviated by making the motion of the piston itself 
open and close the valve. In Fig. 248 the valve is attached 
to the lower end of a rod which passes through a hole in the 
piston. When the piston is pulled up the valve rod is carried 
up with it by friction until it opens the lower valve when its 
upper end strikes against the cylinder cover which checks the 
motion and the piston slides ever the rod. In the downward 
stroke the rod and piston move down together until the valve 
is seated, when the piston again moves along the rod. 



188 

II. Clearance. It is impossible to make the piston fit ac- 
curately against the inside surfaces of the cylinder covers. 
There is always more or less space untraversed by the piston. 
This space is occupied by air which cannot be pumped out. 
An attempt has been made to obviate this difficulty by using 
a mercurial plunger, as in Kravogl's pump. 

III. Air absorbed by the oil used in lubricating the pump. 
This defect cannot be overcome unless the use of oil is abol- 
ished. The mercurial plunger above named and the free-pis- 
ton air pump of Deleuil are attempts in this direction. 

IV. Leakage. There is leakage at the various joints which 
cannot be prevented by the best workmanship. Continual 
inspection and care in the use of the pump are necessary to 
keep it in working order. 

437. — It is obvious that as the exhaustion proceeds the ex- 
ternal unbalanced pressure upon the upper surface of the pis- 
ton approaches nearer to fifteen pounds on each square inch and 
that the work of lifting the piston becomes greater and greater. 
To overcome this difficulty an extra valve is placed in the up- 
per cover of the cylinder which is closed by atmospheric pres- 
sure at the beginning of the down stroke. 

438. — Two cylinders can also be connected to the space to 
be exhausted so that when one piston is at the top of its cyl- 
inder, the other is at the bottom of its cylinder. Such an ar- 
rangement is called a double-barrel air pump. Fig. 248*2. 

Double acting single-barrelled pumps are constructed the 
principle of which becomes evident from an examination of 
Fig. 249, which represents Bianchi's. It will be noticed that 
the upper and lower ends of the cylinder are connected to the 
tube running to the receiver by a bent pipe. The openings 
to the two parts of this pipe may be opened and closed by two 
valves which are attached to the upper and lower ends of a 
rod which passes through a hole in the piston. When the 
piston is pulled up the valve rod is carried with it by friction 
until the upper valve is properly seated, the lower valve being 
opened by the same motion. In the return stroke the rod is 



189 

pulled down, the lower valve is seated and the upper one 
opened the piston itself sliding over the valve rod. The pis- 
ton rod is hollow, the piston valve communicating with the 
interior and the exhausted air passes out through its interior. 
A careful examination will show that the air is exhausted at 
each stroke, while in the ordinary pump a forward and back- 
ward stroke are necessary for the discharge of each barrelful 
of air. 

439. — Sprengel's Pump. In this pump which is used to 
complete the vacuum after the ordinary pump has been used 
up to its capacity, all the preceding defects have been over- 
come. It consists (Fig. 250) of a vertical glass tube about 36 
inches long and 1-10 inch in diameter, open at both ends and 
having a lateral branch near the top. The lower end is in- 
serted in a bottle having a hole in its side. The lower open- 
ing is placed below the level of the outlet so that when in use 
it will be covered by mercury, thus preventing the air from 
rising in the tube. A vessel is placed to receive the mercury 
from the overflow. At the top of the tube a glass funnel is 
attached by means of a rubber tube, around which a spring 
snap is placed. The funnel is filled with mercury and the 
receiver which is to be exhausted is attached to the lateral 
tube. 

Mode of Operation. When the spring-snap is opened the 
mercury in the funnel falls through the vertical tube. The 
air in the receiver expands (§374) and passes into the vertical 
tube in bubbles, which are carried down and escape from the 
lower bottle. As the exhaustion proceeds the bubbles become 
shorter and the fall is accompanied by a sharp click, which 
indicates a good vacuum. When the exhaustion is completed 
no bubbles are visible in the falling column.' 

440. — The Bunsen Pump. This pump, which is very use- 
ful to obtain a partial vacuum, is the same in principle as the 
Sprengel. The falling liquid is water instead of mercury, and 
to obtain the best effect the fall tube should be about 40 feet 
long. When permanently attached to the water pipes of a 
building it is an excellent substitute for the ordinary airpump. 



190 

441- — Richards's exhaust and blast pump (Fig. 251) is a 
convenient laboratory contrivance. 

442. — Applications of the Air Pump. The air pump is 
used in sugar refineries to exhaust the air from the vessel 
containing the boiling syrup. A liquid in a vacuum boils at 
a lower temperature than under atmospheric pressure. 
Burning is thus prevented. Vacuum brakes are used upon 
railroad cars ; pneumatic tubes are arranged for the dispatch 
of packages over considerable distances ; air pumps are em- 
ployed to pump the air from mines and buildings for ventila- 
tion. Fig. 252 represents a pump used in the Harz Moun- 
tains, Germany, for ventilating mines. The same principle 
has been recently employed for the transmission of time. 

MANOMETERS OR PRESSURE GAUGES. 

443. — Gauges are used to measure the pressure either below 
or above that of the atmosphere. 

444. — Vacuum Gauges measure pressures below that of the 
atmosphere. 

445. — 1. The Barometer Gauge (Fig. 253) consists of a 
barometer tube and its cistern. The tube is fixed air tight in 
its cistern by a screw connection and an opening on the side 
of the cistern is in communication with the space the vacuum 
of which is to be measured. As the pressure is removed from 
the surface of the mercury in the cistern the column falls in 
the tube and if the liquid were at the same level in the cistern 
and the tube the vacuum would be complete. 

446. — 2. The Open Tube Gauge. If the top of the barometer 
tube (Fig. 254) be cut off and connection made with the ex- 
hausted space, it is evident that as the vacuum becomes more 
perfect the mercury will be raised in the tube by the atmos- 
pheric pressure on the surface of the mercury in the cistern, 
and if it were complete would stand at a height representing 
the pressure of the atmosphere, 

447. — 3. The Siphon Gauge. Fig. 255 is a modification of 
the barometer gauge. It consists of a U tube, one leg of 
which is filled with mercury. The degree of exhaustion can 
be measured by the difference in level in the two branches. 



i-9 1 

448. — 4. Botirdotfs Metallic Gauge. If the air is exhausted 
from the Bourdon's gauge (Fig. 260) it may be used as a 
vacuum gauge. 

These instruments show the degree of exhaustion and the 
condition of the air pump as to leakage. 

449. — The Condensing" Pump. If the valves of the air 
pump are both reversed air will be forced into the receiver 
instead of being drawn out. When the piston is at the bottom 
of the cylinder, the valves V, and V 2 are both seated. As the 
piston is drawn up its valve V 2 is opened by atmospheric 
pressure above it and the lower valve V, is closed by the ten- 
sion of the air in the receiver. When the piston is forced 
down V, is closed by the compressed air under it and for the 
same reason V, is opened and air is forced into the receiver. 
Every downward stroke of the piston introduces into the con- 
denser a barrelful of air, hence the amount so introduced in- 
creases in an arithmetical ratio. 

450. — Calculation of the compression. Let 6 l be the den- 
sity of the air before compression, V r the volume of the re- 
ceiver and V c the volume of the cylinder and n the number 
of strokes of the piston. Then V r + n V c will be compressed 
to V r and if 6 2 is the final density, by Boyle's Law, 

Vr + nVc :.Vr ::.<?, iff, 

^_^V r |nVc 



Vr 

Also the pressure will be V r -J- n V ( 



, since o, : o 2 :: p 2 : 1. 



V r 

The fact that air is compressed in the condenser may be 
shown by crushing a glass cube, by filling the condenser 
partly full of water and by allowing it to escape through a 
tube which passes into it below the surface of the water ; and by 
the continuous stream due to the air chamber of a forcing 
pump. §265. 

The condensing pump may be modified by rejecting the 
piston valve and using instead a solid plunger. When this 
modification is made a hole must be bored in the side of the 



192 

barrel for the admission of air just below the lower sur- 
face of the piston when the piston is at the end of the stroke. 
(Fig. 256.) 

Another modification is represented in Fig. 257. A lateral 
tube containing a valve opening outward is attached to the 
lower part of the pump barrel. When the plunger is drawn 
up air enters the barrel and when it is pushed down it is forced 
into the lateral chamber. 

The condensing pump is used in a great many important 
operations. The compressed air drill is used in tunneling. 
Compressed air is used in foundries, forges, cupolas, in the 
blow-pipe of chemistry, in blowing glass bottles, in the air 
gun, in ventilating mines and buildings, in the production of 
ice, in the distribution of time, in laying the foundations of 
bridge piers, in sinking spiles, in the brakes of railroad cars, 
in pneumatic dispatch tube, in locomotives. 

Diving bells and dresses, pneumatic call bells, the common 
and organ bellows, gas holders, pneumatic lamps, all illus- 
trate the same principles. Sunken vessels are floated by 
forcing the water out of caissons, attached to them, by com- 
pressed air. Power may be transmitted to considerable dis- 
tances by compressed air. Hero's fountain is a very old 
illustration of its use. §461. 

The degree of compression produced by a condensing 
pump is measured by 

451. — I. The Open Mercurial Pressure-Gauge, (Fig. 258) 
which consists of a long vertical tube fixed in a vessel which 
is connected by a lateral tube to the space subjected to pres- 
sure. Mercury is placed in the cistern. It is obvious that 
when the pressure on the surface equals the atmospheric pres- 
sure in the tube the mercury will be on the same level in both 
tube and cistern ; also that for every 15 pounds tension in the 
enclosed space the height of the mercurial column will be 30 
inches. Hence, if such an instrument were attached to a 
boiler carrying 90 pounds pressure of steam the column would 
have to be 180 inches, or 15 feet high. While for scientific 
purposes this is the best instrument it is impossible to use it 
for ordinary operations. To shorten the tube 



193 

452. — II. The Compressed Air Gauge is used. (Fig. 259.) 
It consists of a U tube closed at one end and connected to the 
space under tension by the other. The bend of the tube is 
filled with mercury. When the tension of the air enclosed in 
the tube is equal to that of the atmosphere the level of the 
mercury in both legs is the same. When the pressure rises 
the enclosed air is compressed and the diminution of volume 
is used to measure the amount of pressure. 

453. — III. Bourdon's Pressure Gauge. This gauge (Fig. 
260) is metallic and consists of a hollow bent tube of elliptic sec- 
tion, closed at one end and having its interior connected with 
the space under tension. To the closed end is attached an in- 
dex which moves over a scale. As the pressure is increased 
the curvature of the tube becomes less, and as it diminishes it 
becomes greater, and these changes are communicated to the 
index by suitable mechanism. The graduation is made by 
comparison with the Open Mercurial Pressure Gauge. 

454. -Water Pumps. I. The Suction Pump is the same as the 
air pump, (§433) in the arrangement of its valves. They both 
open upward. If the connecting pipe of the air pump were 
straightened and placed in a vessel of water it would act like the 
ordinary suction pump. If the level of the water in the suc- 
tion pipe is the same as in the cistern the piston when pulled 
up tends to create a partial vacuum and the air in the pipe expands 
and fills the barrel of the pump ; the tension of air in the pipe 
being diminished the atmospheric pressure upon the water in 
the cistern forces the water up into the pipe to a height cor- 
responding to the diminished tension. When the piston is 
depressed the air will escape through the piston valve V 2 , the 
suction valve V, having closed. Several strokes having been 
made the water will rise to the height of the suction valve 
and above it. At this point in the action if the piston be 
depressed the water will pass through the piston valve and fill 
the barrel above the piston. The water is then lifted up until 
it escapes through the lateral spout. The fact that the water 
is pushed up in the suction pipe by atmospheric pressure may 
be shown by enclosing the cistern under the receiver of the air 



i 9 4 

pump (Fig. 261) and exhausting the air from the surface of 
the water. When so exhausted the pump becomes inopera- 
tive ; if the air is admitted the action is restored. Since, 
therefore, the air pressure causes the rise and (§409) this pres- 
sure can only sustain a column of water 34 feet high, it is ev- 
ident that the suction valve must not be more than that dis- 
tance above the level of the water in the cistern if the water 
is expected to rise to the bottom of the pump barrel. If the 
water is to be pumped out of the lateral pipe this distance 
must be measured from the upper limit of the piston's stroke 
to the surface of the water in the cistern. In practice about 
twenty-eight feet constitute the limit. 

Force required to lift the Piston. Suppose the water 
is at the level of the spout, then the downward pressure on the 
top of the piston equals the pressure of the atmosphere plus the 
area of the piston multiplied by the height of the column of 
water above it or 34 + r^S ; the upward pressure on the bot- 
tom of the piston equals the atmospheric pressure minus the 
weight of the column of water below it or 34 — h 2 S ; hence, 
the difference (34 -f h a S) — (34 — h s S) = hiS + h 2 S=(h 1 -fh a )S 
is the downward pressure of a column of water whose base 
equals the area of piston and whose height is the distance 
from the level of the spout to the level of the water in the cis- 
tern. Hence the same force must be exerted to lift the water 
from the cistern by the pump as if it were lifted in any other 
way. 

If the water is only part of the way up the suction pipe the 
same truth holds, for the downward pressure equals 34 and 
the upward pressure 34 — h s S, h s being the height of the water 
in the pipe ; hence the difference or downward pressure is 
34-( 3 4-h 3 S)-h 3 S. ' 

455.— II. The Forcing Pump. This pump (Fig. 262) usually 
consists of a cylinder having a valve in the bottom opening 
upward and a lateral tube containing a valve opening out- 
ward. A solid plunger works water-tight in the cylinder. 
When the plunger is drawn up the barrel is filled by the air 
pressure and in the down stroke it is forced up through the 



195 

lateral tube. It is evident that the force exerted must be equal 
to the pressure exerted by the weight of a column of water 
whose cross-section is that of the plunger and height the dis- 
tance from the level of the water in the cistern to the top of 
the lateral tube; for suppose the water is forced to the height 
h 2 ; now, as in the suction pump the force necessary to raise 
the piston is l^S ; when the piston is depressed the portion of 
water in the bend is balanced by an equal height in the barrel, 
and force must be applied to lift the column h 2 S. Hence the 
whole force exerted must be h x S + h 2 S, but this is the weight 
of the column above described. 

456. — The Fire Engine represented in Fig. 263 is a double 
forcing pump, the action of which is sufficiently evident from 
the diagram. 

457. — Double Acting Pumps, of which an example is given 
in Fig. 264, lift and force the water at each stroke. 

458. — Rotatory Pumps, in which there is a rotating piston 
upon an axis, have been constructed. (Fig. 265). 

459. — The Air Chamber (Fig. 266) is a large hollow vessel 
attached to the forcing pump to enable the pump to throw a 
continuous stream. It is placed so that the water is forced 
into it through the lateral valve and being filled with air, the 
enclosed air becomes compressed by the inflowing water. 
The water is then forced out by the elasticity of the air. 

460. — Valves are of various forms and are intended to admit, 
cut off or regulate the flow of the liquid. 

461 — Hero's Fountain. Invented by Hero, the philoso- 
pher of Alexandria, 120 B. C, illustrates the production of a 
continuous jet by compressed air. It is a modified form of the 
air chamber. 

A, (Fig. 267) is a glass sphere partly filled with water. A 
small pipe, P,, extends below the level of the water. It is 
evident that if air were forced into the space above the water 
a jet would issue from the pipe and rise to a height propor- 
tioned to the amount of compression. The remainder of the 
apparatus is intended to effect this compression by means of a 



ig6 

column of water. D is a flat vessel containing water, placed 
npon the top of A x ; a pipe P Q extends from the water in this 
basin to the bottom of another sphere, A 2 , below the first. 
From the top of the lower sphere another pipe, P 3 , extends to 
the air space of the first sphere. It is therefore evident that 
if the dish, D, is filled with water, some of it will fall through 
P 2 to A Q , which will gradually become filled, compressing the 
air above it, which escapes through P 3 to the upper sphere A, 
and produces the necessary compression upon the water there. 
The water thus subjected to pressure escapes through the tube 
P r Theoretically, the height of the jet thrown by P x will de- 
pend upon the distance between the levels in D and A 2 . As 
the operation goes on A r becomes empty and A 2 filled. This 
principle having been applied in Hungary to pump water from 
mines, Hero's fountain is sometimes called the Hungarian 
machine. 

462. — The Siphon is a bent tube with one leg longer than 
the other. It is used to draw liquids from vessels. If the 
tube is first filled with the liquid and the shorter leg intro- 
duced into the vessel a stream will issue from the long leg un- 
til the level in the vessel is lowered to the mouth of the 
shorter end. Consider the area of cross section, S, of the tube 
at D. (Fig. 268.) The upward pressure is that of the atmo- 
sphere or 34. The downward pressure is that of the atmo- 
sphere on the surface of the liquid in the vessel, diminished 
by the weight of a column of height B C plus the weight of the 
column BD or 34— BC- S-f-BD - S; hence the difference of pres- 
sure will be 34— B C-S + BD- S— 34=C D S or the weight of 
a column whose length is the distance from the level of the 
water in the vessel to the mouth of the tube. Hence (§417) the 
water will issue with velocity v = 1/2$ • CD. The siphon 
may be filled by immersing it completely in the liquid, or by 
means of the " suction tube." (Fig. 269.) The Wurtemburg 
siphon (Fig. 270) is a tube bent twice at right angles, the ob- 
ject being to prevent the escape of liquid and thus avoid re- 
peated fillings. 

463. — The Intermitting Siphon (Fig. 271) is a siphon 
placed in a vessel which is filled with water by a tube of 



i 9 7 

smaller diameter. When the level of the water reaches the 
highest point of the siphon the tube will be filled and the ves- 
sel will be emptied faster than the supply tube can fill it. 
When the level reaches the mouth of the short tube of the 
siphon air will enter and the siphon will become inoperative 
until the vessel is again filled. 

464. — Intermittent Springs (Fig. 272) are explained on this 
principle. 

Lampwick and woven goods act as siphons when hung 
over the edge of a vessel with the short length in the liquid. 
Tall chimneys used for ventilation are equivalent to inverted 
siphons. 

465. — The Bellows. — I. The simple bellows (Fig. 273), con- 
sists of two triangular shaped boards held together by a flexi- 
ble border of leather fastened to their edges. A tube is 
attached to the apex of the box thus formed. A valve in the 
centre of the lower board opens inward. When the boards 
are separated the upward pressure of the air opens the valve 
and rushes inside to fill the partial vacuum thus formed. 
When the boards are brought together the valve is closed by 
its own weight and by the compressed air which escapes 
through the discharge pipe. It is evident that the action is 
intermittent, the air escaping in jets. 

466. — II. The compound bellows (Fig. 274) is constructed to 
overcome this defect. If the lower board of the simple bel- 
lows be fixed and another simple bellows added to it below, as 
is represented in the diagram, and a weight placed upon the 
upper board a continuous jet will be delivered when the lever 
is worked. It is seen that the air instead of passing through 
the blast pipe immediately, enters the upper chamber, where 
it is compressed by the load placed upon the top and then 
escapes through the blast pipe. This part of the apparatus 
resembles the air-chamber of the forcing pump. §459. 

467. — The Pipette (Fig. 275) consists of a glass tube drawn 
to a small point. When the lower end of the tube is intro- 
duced into a liquid, by reason of hydrostatic pressure it will 



rise to the same height in the tube as the level is outside. If 
the upper end be now closed with the finger and the tube be 
removed the liquid will remain in the tube ; if the finger is 
removed the liquid will fall into any vessel placed under it. 

468. — Mariotte's Bottle (Fig. 276) is used (1) to give a con- 
stant discharge of liquid ; (2) to drive gas from a vessel with 
uniform velocity ; (3) to draw gas with accelerated velocity. 
It consists of a bottle having a tubulure in the side near the 
bottom. A tube open at both ends passes through a cork 
which fits tight in the neck of the bottle. 

1. Let the tube and bottle be completely filled with water 
and the tube reach a position a above the outlet of the bottle. 
The pressure inward upon the liquid in the tubulure is due 
to the atmospheric pressure or 34 feet ; the pressure outward 
upon the same molecules of water is the pressure of the atmo- 
sphere plus the weight of a column of water whose height 
extends from the tubulure to the level of the water in the 
tube, or 34 4- h x S. Hence, there is an outward pressure of 
(34 + h,S) — 34 = hjS ; hence if the outlet is opened water 
will flow. The level will fall in the tube as the water escapes 
from the side of the bottle and the velocity will diminish as 
may be seen by the diminishing range. 

2. Suppose the water has fallen in the vertical tube to the 
level a. The water will still flow from the tubulure while 
air enters through the tube in bubbles which rise to the upper 
part of the bottle, in which case the tension of the enclosed 
air plus the weight of the column of water above a equals the 
atmospheric pressure in the tube, and the velocity of efflux 
from the tubulure is due to the weight of the column of water 
extending from the tubulure to the plane a or h 2 S. As this 
column is constant the velocity will be constant. Hence the 
bottle may be used for obtaining a constant discharge of 
liquid. 

If the water as it escapes is allowed to fall into a vessel con- 
taining gas, as the water flows in uniformly the gas will be 
forced out uniformly. 



Gas may also be drawn into the bottle by attaching the ver- 
tical tube to a gas receiver. As the surface of the water de- 
scends uniformly, the density of the gas continually increases 
and the flow is uniformly accelerated. 

It is evident that the tube may be withdrawn from the bot- 
tle and an extra tubulure placed in the side with the same re- 
sult. This form of bottle is represented in Fig. 277. 

469. — The Magic Funnel, (Fig. 278) illustrates the upward 
pressure of the air. It is double ; near the handle is a hole 
over which the finger may be placed. The space between the 
two funnels can be filled with water, which is kept there by 
atmospheric pressure. Water may be run through the inside 
funnel and colored liquid may be caused to fall out of the annu- 
lar opening in the neck by removing the finger from the vent. 

470. — The Hydraulic Ram illustrates the inertia of water. 
(§258). Fig. 279 represents an inclined tube connected with a 
reservoir. At the end of this tube is a valve opening down- 
ward. A short distance from the end is an air chamber with a 
valve opening upward. A vertical discharge pipe is taken 
from near the bottom of the air chamber to the desired 
vertical height to which it is intended to pump the water. 

Mode of Operation. The water rushes down the inclined* 
pipe and escapes from the open valve ; after the water has 
gained sufficient momentum it closes the valve. When the 
valve is closed and the escape of the water cut off the recoil 
lifts the air chamber valve. The water enters, compresses the 
air and rises in the delivery pipe. When the momentum is 
expended the valve in the air chamber closes, the escape valve 
opens and the same operation is repeated. A very large quan- 
tity of water is used to force a comparatively small quantity 
to a height. 

471. — The fall of water may be used as a motive power. 
Wheels with horizontal 'and vertical axles have been employed 
for this purpose. Among the first class are 

472.— I. Wheels with Horizontal Axles. 1. The Overshot 
wheel (Fig. 280) receives the water in buckets arranged 



200 

on its circumference. The weight of the water acting at the 
end of an arm varying from zero to the horizontal radius 
upon one side of the wheel produces rotation. The moment 
of this weight increases from the top of the wheel to the end 
of the horizontal radius, after which it diminishes from short- 
ness of arm and loss of water until it is nothing at the bottom 
of the wheel. 

473. — 2. The Undershot wheel (Fig. 281) receives its mo- 
tion from the impulse of water against radial floats placed 
upon its periphery. 

474. — 3. The Breast wheel (Fig. 282) is placed in a curved 
channel called u the breast" and receives the water upon the 
level of its axis. The water acts partly by weight and partly 
by impulse. 

475. — II. Wheels with Vertical Axles, r. Barker's Mill, 
(Fig. 283) an early form of the reaction wheel, consists of a vertical 
hollow tube having hollow horizontal arms at its lower end. 
Openings are made in these arms near the ends on opposite 
sides. The tubes are filled with water which escapes from 
the orifices. The hydrostatic pressure on one side being used 
to produce the flow, the pressure on the other side is unbal- 
anced and produces motion of the mill opposite to the direc- 
tion of the jets. 

476. — 2. The Turbine Wheel consists of a drum containing 
vanes which are bent back upon themselves, and a drum with 
guide blades which form curved partitions through which the 
water is directed upon the vanes of the wheel. The water 
before entering and after leaving the wheel may flow parallel 
to the vertical axis (Fig. 284) in which case the wheel is called 
a parallel flozv turbine ; it may flow outward from the axis or 
inward towards the axis. In the former case it is called an 
outward flow turbine (Fig. 285), and in the latter an inward 
flow turbine (Fig. 286). These wheels are small for their 
power and equally efficient for great and small heads of 
water. 



PROBLEMS, 



Section 15. 

1. If a line 15 inches long represent a force of 6, what force 
will a line 2 feet long represent ? Ans. 

2. A line 1 inch in length represents 50 lbs. What length of 
line will represent one ton (2240 lbs.) ? Ans. 

3. A force of ¥ 1 lbs. is represented by a line a inches long, what 
force will be represented by a line b inches long ? Ans. 

4. A line a inches long represents ¥ 1 lbs. F 2 lbs. would be rep- 
resented by a line how long? Ans. 

5. A string is attached at one end to a ceiling ; at the other end 
is a 6 lb. weight and in the middle a 2 lb. weight. What are the 
tensions on the different parts of the string ? Ans. 

Sections 17 and 18. 

1. Two forces F x =6 and F 2 =10 make an angle of 30° with each 
other. What is the magnitude and direction of the resultant ? Ans. 

2. A force F t — 6 and the resultant of it and a second force, F 2 , 
is 14. The resultant makes an angle of 10° with F r What is the 
magnitude and direction of the second force ? Ans. 

3. A force of 10 has two components which make angles of 10° 
and 50° with it. What are the components ? Ans. 

4. The resultant of two forces is at right angles to one of the 
forces. Show that it is less than the other force. 

5. The resultant of two forces is at right angles to one force and 
is equal to half the other. Compare the forces. 

6. Two forces act upon a particle — the first has twice the mag- 
nitude of the second — add 6 to the larger and double the smaller ; 
the direction of the resultant is unchanged. What are the forces? 

7. Show that if the angle between two forces acting upon a par- 
ticle is increased the resultant is diminished, and vice versa. 

8. The components of a force are 10 and 20. The force 10 
makes an angle of 10° with it. Find the magnitude and direction 
of the force. 

9. How can the forces 25 and 75 be applied to a particle so that 
their resultant may be 60 ? 



10. Find the magnitude of the resultant of two forces, 10 kilo- 
grammes and 8 kilogrammes, which act at an angle of 105°. 

11. Two forces F^IS and F 2 =25 are at right angles to each 
other. What is the magnitude and direction of the resultant ? 

11. Two forces whose magnitudes are as 3 : 4 liave a resultant of 
15 when at right angles to each other. What are the forces ? 

Section 22. 

1. Find the resultant in magnitude, direction and point of ap- 
plication of two forces -4-6, — 6 in the same straight line. 

2. Of +6, -f-6. 

3. Of -f-6, -f-6, -f- 12, — 28, and explain the meaning of the minus 
sign. 

4. Show T that three forces, 1,2, 3, in order to keep a body at rest 
must act in the same straight line. 

5. Two forces acting upon a particle have greatest and least re- 
sultants of 72 and 56. What are the forces? 

Section 23. 

1. Three forces, 6, 8, 10, make angles of 30° and 40° with each 
other. Find their resultant. 

2. Five forces are applied to a point A. The second makes an 
angle with the first of 20°, the third 19° with the second, the fourth 
21°30 / with the third, and the last 25° with the fourth. The first, 
second and third are equal, the other forces one half greater. Re- 
quired the magnitude and direction of the resultant ? 

3. A particle is urged by three equal forces. The first makes an 
angle of 90° with the second, and the third an angle of 135° with 
the first. Required the magnitude and direction of the resultant ? 

4. A O B and COD are chords of a circle which intersect at 
right angles at O ; forces are represented in magnitude and direc- 
tion by OA, OB, OC, OD ; show that their resultant is represented 
in direction by the straight line which joins O to the center of the 
circle, and in magnitude by twice this straight line. 

5. Eight points are taken on the circumference of a circle at 
equal distances, and from one of the points straight lines are drawn 
to the rest. If these straight lines represent forces acting at the 
poipt, show that the direction of the resultant coincides with the 
diameter through the point, and that its magnitude is four times 
that diameter. 



6. The circumference of a circle is divided into a given even 
number of equal parts, and from one of the points of division 
straight lines are drawn to the rest. Show that the direction of the 
resultant coincides with the diameter through the point. 

7. Forces of 3, 4, 5, 6 lbs. respectively act along the straight 
lines drawn from the centre of a square to the angular points taken 
in order. Find their resultant. 

8. Three forces represented by the numbers 1, 2, 3, act on a par- 
ticle in directions parallel to the sides of an equilateral triangle 
taken in order. Determine their resultant. 

Sections 24 and 25. 

1. Given two parallel forces -f-10 and -{-20, find the magnitude, 
direction, point of application and sign of the resultant ? 

2. The forces are +10 and -flO. Find the resultant. 

3. The forces are +10 and — 20. Find the resultant. 

4. The forces are — 20 and — 15. Find the resultant. 

5. The forces are — 10 and +10. Find the resultant. 

6. Two men carry a weight of 152 lbs. between them on a pole, 
resting on one shoulder of each ; the weight is three times as far 
from one man as from the other. Find how much weight each 
supports, the weight of the pole being disregarded. 

7. A man supports two weights strung on the ends of a stick 40 
inches long placed across his shoulder ; if one weight be two-thirds 
of the other, find the point of support, the weight of the stick being 
disregarded. 

8. Parallel forces are applied at two points 5 inches apart, and 
are kept in equilibrium by a third force 3 inches from the one and 
2 inches from the other. What is the ratio of the forces ? Ans. 2 : 3. 

9. A beam, the weight of which is equivalent to a force of 10 
pounds acting at its middle point, is supported on two props at the 
end of the beam. If the length of the beam be 5 feet, find where 
a weight of 30 pounds must be placed so that the pressure on the 
two props may be 15 pounds and 25 pounds respectively. Ans. 20 in. 

10. Two weights of 3 ounces and 5 ounces hang at the ends of a 
rod 12 inches long, and a third weight of 6 ounces is placed 3 inches 
from the lighter weight ; find the position of the resultant. 

Ans. 6 3-7 inches from one end. 

11. The ratio of two unlike parallel forces is four-fifths, and the 



distance between them is 10 inches ; find the position of the re- 
sultant. Ans. 40 inches from greater force. 

12. The resultant of two unlike forces is 6 pounds and acts 8 
inches from the greater force, which is 10 pounds ; find the dis- 
tance between them. Ans. 12 inches. 

13. Forces of 3, 4, 5, 6 pounds act at distances of 3 inches, 4 
inches, 5 inches, 6 inches from the end of a rod ; at what distance 
from the same end does the resultant act. Ans. 4 7-9 inches 

14. If a weight rest in the middle of a square rough table, will 
the pressure on each leg be altered if one pair of legs be longer than 
the opposite pair ? 

15. Find the centre of like parallel forces, 3, 2, 5, 7 pounds, 
which act at equal distances apart along a straight rod 12 inches 
long. Ans. 7 13-17 inches from 3 pounds weight. 

16. Forces 6, 8, 10, — 12 and — 16 act on a rod at right angles to 
it, the distance between the lines of action of 6 and 8 is 4 feet ; be 
tween 8 and 10 two feet ; between 10 and — 12 fourteen feet and 
between — 12 and — 16 seven feet. Required the magnitude, di- 
rection, sign and point of application of the resultant. 

17. Forces — 10, -flO, +20, — 20 are parallel. Find their re- 
sultant. 

18. Forces 1, — 2, — 3, 4 act on a rod at right angles to it, and at 
equal distances in the same plane. Find their resultant. 

19. Forces F, — 3F, — 5F, 7F act on a rod at distances from one 
end proportional to their respective magnitudes ; the forces are 
parallel and in the same plane. Find their resultant. 

Section 37. 

1. Two equal and weightless rods are jointed together and form 
a right angle. They move freely about their common point. Find 
the rates of the weights that must be suspended from their extrem- 
ities that one of them may be inclined to the horizon at 60°. 

Ans. 1 : i/~s, 

2. A rod A B moves about a fixed point B. Its weight W acts 
at its middle point, and it is kept horizontal by a string A C that 
makes an angle of 45° with it. Find the tension in the string. 

Ans. — — - 
V 2 



3. A beam 3 feet long, the weight of which is 10 pounds, and 
acts at its middle point, rests on a rail, with 4 pounds hanging 
from one end and 13 pounds from the other ; find the point at 
which the beam is supported ; and if the weights at the two ends 
change places, what weight must be added to the lighter to pre- 
serve equilibrium ? Ans. 12 inches ; 2T pounds. 

4. Show that the sum of the moments of two forces represented 
by the two sides of a triangle about any point in the base is con- 
stant. 

5. A rod 10 inches long can turn freely about one of its ends ; 
a weight of four pounds is hung to a point 3 inches from this end ; 
what vertical force at the other end is required to support it ? 

Ans. 1.2 pounds. 

6. Three forces represented in magnitude, direction and posi- 
tion, by the sides of a triangle, produce a couple. 

1. Several couples in a plane whose forces are parallel, are ap- 
plied to a rigid right line ; required the resultant couple. 

8. Several couples in a plane, whose respective arms are not par- 
allel, act upon a rigid right line ; required the resultant couple. 

Section 50. 

1. A weight W (Fig. 1) is attached to a string, which is secured 
at A, and is pushed from a vertical by a strut C B ; required the 
pressure F on B C when the angle C A B is 0. 

Ans. F =W tan. 0. 

2. A brace, AB (Fig. 2) rests against a vertical wall and upon 
a horizontal plane, and supports a weight W at its upper end ; re- 
quired the compression upon the brace and the thrust at A when 
the angle C A B is 0. Ans. F = W sec. 0. 

3. A rod (Fig. 3) whose length is B C — 1 is secured at a point 
B, in a horizontal plane, and the end C is held up by a cord A C so 
that the angle AB C is 0, and the distance A B =a ; required the 
tension on AC and compression on B C, due to a weight W applied 

at C * Ans. Tension = I W cot. 0. 

a 

4. A cord whose length AC =1 is secured at two points in a 
horizontal line, and a weight W is suspended from it at B ; re- 
quired the tension on each part of the cord. 



Section 59. 

1. A resultant of 153 toward the north is produced by the forces 
100 and 125 ; how are they inclined to the meridian ? 

2. A resultant of 617 divides the angle between its components 
into 28° and 74° ; what are the components ? 

3. Explain how a vessel is enabled to sail in a direction nearly 
opposite to that of the wind. 

4. Explain bow the force of the current may be taken advan- 
tage of to urge a ferry-boat across a river ; the center of the boat 
being attached, by means of a long rope, to a mooring in the mid- 
dle of the stream. 

5. Show why the traces of a horse oaght always to be parallel 
to the road along which he is pulling. 

6. Indicate the forces that maintain a kite in equilibrium. 

Section 69. 

Solve problems of Section 23 by the method of rectangular co- 
ordinates. 

Section 76. 

1. Four vertical forces of 4, 6, 7, 9 pounds act at the four 
corners of a square ; find their resultant. 

Ans. 5-13 of middle line from one of the sides. 

2. A flat board 12 inches square is suspended in a horizontal 
position by strings attached to its four corners ABCD, and a 
weight equal to the weight of the board is laid upon it at a point 
3 inches distant from the side AB and 4 inches from AD ; find 
the relative tensions in the 4 strings. 

Ans. As. I : 1 : - 1 « 5 

3. Required the point of application of the resultant of three 
equal weights, applied at the vertices of a plane triangle. 

4. Required the point of application of the resultant of a system 
of equal parallel forces, applied at the vertices of a regular polygon. 

Ans. At the center of the polygon. 

5. Parallel forces of 3, 4, 5 and 6 pounds are applied at the suc- 
cessive vertices of a square, whose side is 12 inches. At what dis- 
tance from the first vertex is the point of application of their resul- 
tant ? Ans. 9.475 inches 



6. Seven equal forces are applied at seven of the vertices of a 
cube. What is the distance of the point of application of their 
resultant from the eighth vertex? 

Section 86. 

1. A particle is kept at rest by three inclined forces ; what is 
the value of the third force when the other two have the same line 
of action and are opposite in direction to each other. 

2. In the above problem what would be the value of either one 
of the forces whose lines of action are common when the third force 
has magnitude. 

3. If three forces represented by the numbers 1, 2, 3, acting in 
plane keep a particle at rest, how must they act ? 

4. Can forces 3, 4, 7 keep a particle at rest when they act in one 
plane ? 

5. Three forces at a point are parallel to the sides of a triangle 
and are inversely as the perpendiculars from the angular points of 
triangle, on the sides parallel to which the forces act respectively. 
What is their resultant ? 

6. Two forces of 5 and 3 act upon a particle. The angle be- 
tween the two forces is 90°. What third force must be applied to 
produce balance. 

f. The circumference of a circle is divided into any number of 
equal parts ; forces are represented in magnitude and direction by 
straight lines drawn from the center to the points of division : show 
that these forces are in equilibrium. 

8. A weight of 25 pounds hangs at rest, attached to the ends of 
two strings, the lengths of which are 3 and 4 feet, and the other 
ends of the strings are fastened at two points in a horizontal line 
distant 5 feet from each other : find the tension of each string. 

Section 93. 

1. Parallel forces are applied at two points five inches apart, 
and are kept in equilibrium by a third force three inches from the 
one and two inches from the other. What is the ratio of the forces ? 

2. Equal weights hang from the corners of a triangle ; find the 
point at which it must be suspended to rest horizontally. 

Section 108, &c. 

1. Find the common centre of gravity of three equal bodies 
placed at the three angles of any triangle. 



' 8 

2. At the three angles of the triangle ABC, whose sides are as 
follows, viz. : A B 5, B 4 and A C 2, are placed three weights, 
viz : 3 pounds at A, 2 pounds at B and 1 pound at C ; required their 
centre of gravity. 

3. Find the centre of gravity of 5 equal bodies placed at five of 
the angles of a regular hexagon. 

4. What force applied at the centre of gravity of a barrel of 
flour, standing upon one end, will be sufficient to overturn it, if its 
weight be 200 pounds, its length 29 inches and the diameters of its 
ends 18 inches? And what will be the pressure upon the floor 
when it begins to move? 

5. How far may the aforesaid barrel be inclined from its per- 
pendicular position before it will fall ; and what will then be its 
pressure upon the floor ? 

6. Find the minimnm inclination of a plane to the horizon at 
which a cube of uniform density will roll down it. 

7. Find the same for a pentagonal prism. 

8. A weight of 100 pounds is suspended by a cord. How far 
will a force of 30 pounds draw it from its vertical position ; and 
what, in that position, will be the pressure upon the point of sus- 
pension ? 

9. A weight of 500 pounds is suspended by a cord which is capa- 
ble of sustaining a weight of only 100 pounds. How far will it 
descend from the level of the point of suspension before the cord 
will break ? 

Section 130. 

1. A weight of 5 pounds hung from one extremity of a straight 
lever balances a weight of 15 pounds hung from the other : find the 
ratio of the arms. 

2. Two weights of 3^- pounds and 4^ pounds are hung at the 
ends of a straight lever whose length is 92 inches : find where the 
fulcrum must be for equilibrium. 

3. Two weights which together weigh 6j pounds are hung at 
the ends of a straight lever and balance : if the fulcrum is four 
times as far from one end as from the other find each weight. 

4. A lever 7 feet long is supported in a horizontal position by 
props placed at its extremities : find where a weight of 28 pounds 
must be placed so that the pressure on one of the props may be 
eight pounds. 



5. Two weights of 12 pounds aud 8 pounds respectively at the 
ends of a horizontal lever 10 feet long balance ; find how far the 
fulcrum ought to be moved for the weights to balance when each is 
increased by 2 pounds. 

6. If the pressure on the fulcrum be equivalent to a weight of 
15 pounds, and the difference of the forces to a weight of 3 pounds, 
find the forces and the ratio of the arms at which they act. 

7. A lever is in equilibrium under the action of the forces P and 
Q, and is also in equilibrium when P is trebled and Q is increased 
by 6 pounds ; find the magnitude of Q. 

8. The pressure on the fulcrum is 12 pounds, and the distance of 
the fulcrum from the middle point of the lever is one-twelfth of the 
whole length of the lever ; find the forces which acting on oppo- 
site sides of the fulcrum will produce equilibrium. 

9. One force is four times as great as the other, and the forces 
are on the same side of the fulcrum, and the pressure is 9 pounds on 
it ; find the position and the magnitude of the forces. 

10. A B C is a straight weightless rod 9 inches long, placed be- 
tween two pegs A and B which are 4 inches apart, so as to be kept 
horizontal by means of them and a weight of 10 pounds hanging at 
C ; find the pressures on the pegs. 

11. A lever bent at right angles, with the angle for fulcrum, and 
having one arm double the other, has two weights hanging from its 
ends ; if in the position of equilibrium the arms are equally in- 
clined to the horizon, compare the weights. 

12. A straight uniform lever whose weight is 50 pounds and length 
6 feet, rests in equilibrium on a fulcrum when a weight of 10 pounds 
is suspended from one extremity ; find the position of the fulcrum 
and the pressure on it. Ans. 2 -J- feet from the end at which 10 
pounds is suspended ; 60 pounds. 

13. Two weights P and Q hang at the ends of a straight heavy 
lever whose fulcrum is at the middle point ; if the arms are both 
uniform, but not of the same weight, and the system be in equilib- 
rium, show that the difference between the weights of the arms 
equals twice the difference between P and Q. 

14. A bar of iron of uniform section and 12 feet long is supported 
by two men, one of whom is placed at one end; find where the other 






IO 

must be placed so that be may sustain tbree-fiftbs of the' whole 
weight. Ans. Two feet from the other end. 

15. Two weights of 2 pounds and 5 pounds balance on a uniform 
heavy lever, the arms being in the ratio of 2 to 1 ; find the weight 
of the lever. Ans. 2 pounds. 

16. A straight lever weighing 20 pounds is moveable about a ful- 
crum ata distance from one extremity equal to one-fourth of its 
length ; find what weight must be suspended from that extremity 
in order that the lever may remain at rest in all positions. 

Ans. 20 pounds. 

17. A bent lever is composed of two straight uniform rods of the 
same length, inclined to each other at 120°, and the fulcrum is at 
the point of intersection ; if the weight of one rod be double that 
of the other, show that the lever will remain at rest with the lighter 
arm horizontal. 

18. Two men carry a uniform plank 6 feet in length and weigh- 
ing 2 cwt., and at 2 feet from one end a weight of 1 cwt. is placed; 
if one man have this eud resting on his shoulder, find where the 
other man must support the beam in order that they may share the 
whole weight equally. Ans. 8 inches from the other end. 

Section 136. 

1. A body, the weight of which is one pound, when placed in one 
scale of a false balance appears to weigh fourteen ounces ; find its 
weight when placed in the other scale. Ans. 18 2-7 ounces. 

2. The arms of a balance are in the ratio of ] 9 to 20 ; the pan 
in which the weights are placed is suspended from the longer arm ; 
find the real weight of a body which apparently weighs 38 pounds. 

Ans. 40 pounds. 

3. If a balance be false, having its arms in the ratio of 15 to 16, 
find how much per pound a customer really pays for tea which is 
sold to him from the longer arm at 90 cents per pound.. Ans. 

Section 138. 

1. The movable weight for which the steelyard is constructed is 
one pound, and a tradesman substitutes a weight of two pounds, 
using the same graduations, thus giving his customers a weight 
which he says is the same number of times the movable weight as 
before ; show that he defrauds his customers if the centre of grav- 



II 

ity of the steelyard be in the longer arm, and himself if it be in the 
shorter arm. 

2. The movable weight is one pound and the weight of the beam 
is one pound ; the distance of the point of suspension from the 
body weighed is 2-J- inches, and the distance of the centre of grav- 
ity of the beam from the body weighed is 3 inches ; find where the 
movable weight must be placed when a body of 3 pounds is weighed. 

Ans. 1 inches from the point of suspension. 

3. If the fulcrum divide the beam, supposed uniform, in the ra- 
tio of 3 to 1, and the weight of the beam be equal to the movable 
weight, show that the greatest weight which can be weighed is four 
times the movable weight. 

1 

4. If the beam be uniform and its weight — of the movable 

weight, and the fulcrum be - of the length of the beam from one 

end ; show that the greatest weight which can be weighed is 

2m (n — l)-f-n-}-2 . , ,, . , 
> LA ! — times the movable weight. 

2m 

5. Find what effect is produced on the graduations by increas- 
ing the movable weight. Ans. The point from which the gradua- 
tions begin, is brought nearer to the point of suspension. 

6. Find what effect is produced on the graduations by increas- 
ing the density of the material of the beam. Ans. The point from 
w T hich the graduations begin is taken further from the point of sus- 
pension. 

Seetion 140. 

1. Find the radius of the wheel to enable a power of \\ pounds 
to support a weight of 28 pounds, the diameter of the axle being 6 
inches. Ans. 56 inches. 

2. Find what weight suspended from the axle can be supported 
by 3 pounds, suspended from the wheel, if the radius of the axle is 
1-J- feet, and the radius of the wheel is 3^ feet. Ans. <6\ pounds. 

3. If the radius of the wheel be to the radius of the axle as 8 is 
to 3, and two weights of 6 pounds and 15 pounds respectively be 
suspended from the circumferences of the wheel and axle, find what 
weight will descend. Supposing that the weight which tends to de- 



12 

scend is supported by a prop, find the pressure on the prop and on 
the fixed supports of the wheel and axle. Ans. The weight of 6 
pounds. The prop must support f pound, leaving 5f pounds on the 
wheel to balance the 15 pounds on the axle. The pressure on the 
fixed supports is 2 Of pounds. 

4. The radius of the axle of a capstan is 2 feet, and six men push 
each with a force of one cwt. on spokes 5 feet long ; find the tension 
they will be able to produce in the rope which leaves the axle. 

Ans. 15 cwt. 

5. The radius of the wheel being three times that of the axle, 
and the string on the wheel being only strong enough to support a 
tension of 36 pounds, find the greatest weight which can be raised. 

Ans. 108 pounds. 

6. In the wheel and axle if the two ropes are coiled each on 
itself, and their thickness not neglected, find whether the ratio of 
the power to the weight would be increased or diminished as the 
weight was raised, supposing the ropes to be of the same thickness. 

Ans. Increased. 

Section 142. 

1. In a single movable pulley if the weight of the pulley be 2 
pounds, find the force required to raise a weight of 4 pounds. 

Ans. 3 pounds. 

2. If there be two strings at right angles to each other and a 
single movable pulley, find the force which will support a weight of 

j/ 2 pounds. Ans. One pound. 

3. A man stands in a scale attached to a movable pulley, and a 
rope having one end fixed passes under the pulley, and then over a 
fixed pulley ; find with what force the man must hold down the free 
end in order to support himself, the strings being parallel. 

Ans. A force equal to a third of his weight. 

Section 143. Cop. 1. 

1. If the weight be represented by the height of the plane, show 
what straight line represents the pressure on the plane. 

Ans. The perpendicular from the right angle on the length. 

2. If F 4 = 12 pounds, and the height of the plane be to its base 
as 3 is to 4, find F,. Ans. 11 pounds. 



J 3 

3. If F 2 = 10 pounds and F l = 6 pounds, find R. Ans. 8 lbs. 

4. If Fj = R, find the inclination of the plane, and the ratio of 
F 1 to F 2 . Ans. 45° :: as 1 is to |/^~ 

5. If F x is to R as 3 is to 4, express each of them in terms of 
F 2 . Ans. F l = 3-5 F 2 , R = 4-5 F 2 . 

6. If Fj = 9 pounds, find F 9 , when the height of the plane is 3 
inches, and the base 4 inches. Ans. 15 pounds. 

7. An inclined plane rises 3 feet 6 inches for every 5 feet of 
length. If F 2 = 200, find F x . Ans. 140 pounds- 

8. A railway train weighing 30 tons is drawn up an inclined 
plane of 1 foot in 60 by means of a rope and a stationary engine. 
Find what number of pounds at least the rope should be able to 
support. Ans. 1120. 

9. A weight of 20 pounds is supported by a string fastened to a 
point in an inclined plane, and the string is only just strong enough 
to support a weight of 10 pounds. The inclination of the plane to 
the horizon being gradually increased, find when the string will 
break. Ans. At an inclination of 30°. 

Section 143. Cor. 1. 

1 . If F 2 = 12 pounds, and the base be to the length as 4 is to 5, 
find F r Ans. 9 pounds. 

2. If F 2 = 48 pounds, and the base be to the height as 24 is to 
1, find F 1 and R. Ans. 14 pounds ; 50 pounds. 

3. If R = 2 pounds and F 2 = 1 pound, find F 2 , and the inclina- 
tion of the plane. Ans. |/~3~pounds; 30°. 

4. If F 2 = 12 pounds, and F, = 9 pounds, find R. 

Ans. 15 pounds. 

Section 144. 

1. In a screw which has seven threads to the inch, find the pres- 
sure that can be produced by a force of 6 pounds applied at the 
circumference, the radius of the cylinder being one inch. 

Ans. 264 pounds. 

2. If the circumference of a screw be 10 inches, what force must 
be applied to overcome a resistance of 30 pounds, the distance be- 
tween the threads being one-fourth inch ? Ans. 12 ounces. 



3. How many turns must be given to a screw formed upon a 
cylinder whose length is 10 inches, and circumference 5 inches, that 
a power of 2 ounces may overcome a pressure of 100 ounces ? 

Ans. 100. 

4. A screw is made to revolve by a force of 2 pounds applied at 
the end of a lever 3.5 feet long ; if the distance between the threads 
be one half inch, what pressure can be produced ? 

Ans. 9 cwts., 1 qr., 20 lbs. 

5. If a power of 1 pound describe a revolution of 3 feet, whilst 
the screw moves through one-fourth inch, what pressure will be 
produced ? Ans. 1 cwt., 1 qr., 4 lbs. 

6. The circumference described by the power is 4 feet, and the 
distance between the threads is one-third inch ; what power is re- 
quired to produce a pressure of 1 ton ? Ans. 15 5-9 pounds. 

7. The tangent of the angle of a screw is J, the radius of the 
cylinder 4 inches, and the length of the power-arm 2 feet. Find 
the ratio of F 2 to F 2 . Ans. 24 : 1. 

8. A screw having a head 12 inches in circumference is so formed 
that its head advances a quarter of an inch at each revolution ; find 
what force must be applied at the circumference of the head that 
the screw may produce a pressure of 96 pounds. Ans. 2 pounds. 

Section 146. 

1. A wedge is right-angled and isosceles, and a force of 50 
pounds acts opposite to the right angle. Determine the other two 
forces. Ans. 25, j/2 pounds. 

2. A wedge is in the form of a equilateral triangle, and two of 
the forces are 40 pounds each ; find the third force. 

Ans. 40 pounds. 

3. Find the vertical angle of an isosceles wedge when the pres- 
sure on the face opposite this angle is equal to half the sum of the 
two resistances. Ans. 60°. 

Section 147. 

1. Three horizontal levers AJAA3, A 3 A / A 4 , A 4 A // A 5 without 
weight, whose fulcrums are A, A 7 , A", act on one another at A 3 and 
A 4 respectively, and are kept in equilibrium by weights of 1 pound 
at A % and 24 pounds at A 5 ; if A,A, AA 3 , A 3 A 4 , A 4 A // , A"A 5 are 



*5 

equal to 2, 1, 4, 4, 1 feet respectively, find the position of A/ and 
the pressure on it. Ans. A 3 A / = 3 feet ; 8 pounds. 

2. In a combination of wheels and axles each of the radii of the 
wheels is five times the radius of the corresponding axle. If there 
be three wheels and axles determine what power will balance a 
weight of 375 pounds. Ans. 3 pounds. 

3. A rope, the ends of which are held by two men A and B, 
passes over a fixed pulley L, under a movable pulley M, and over 
another fixed pulley N. A weight of 120 pounds is suspended from 
M. Supposing the different parts of the rope to be parallel find 
with what force A and B must pull to support the weight. 

Ans. 60 pounds. 

4. A is a fixed pulley ; B and C are movable pulleys. A string 
is put over A ; one end of it passes under C and is fastened to the 
centre of B ; the other end passes under B and is fastened to the 
centre of A. Compare the weights of B and C that the system 
may be in equilibrium, the strings being parallel. 

Ans. The weight of C is twice the weight of B. 

5. Two unequal weights connected by a fine string are placed 
on two smooth inclined planes which have a common height, the 
string passing over the intersection of the planes, find the ratio be- 
tween the weights when there is equilibrium. 

Ans. The weights are as the lengths of the planes on which 
they are placed. 

6. A weight W is supported on an inclined plane by a string 
along the plane. The string passes over a fixed pulley, and then 
under a movable pulley to which a weight W is attached, and hav- 
ing the parts of the string on each side of it parallel ; the end of 
the spring is attached to a fixed point. Show that in order that the 
system may be in equilibrium the height of the plane must be half 
its length. 

Section 171. 

1. What is the mechanical advantage of a wheel and axle in 
which the diameter of the axle is 3 inches, and the radius of the 
wheel 12 inches ? Ans. 8. 

2. Is the mechanical advantage of a wheel and axle increased or 
diminished by lessening the radii of wheel and axle by the same 
amount? 



16 

3. Find the mechanical advantage of a system of three movea- 
ble pulleys arranged as in Fig. — . 

4. The length of the power-arm is 15 inches ; find the distance 
between two consecutive threads of the screw, that the mechanical 
advantage may be 30. Ans. it inches. 

5. If a screw makes m turns in a cylinder a foot long, and the 
length of the power-arm is n feet, find the mechanical advantage. 

Ans. 2 tz m n. 

6. If on a wheel and axle the mechanical advantage be six times 
as great as on a single movable pulley, compare the radii of the 
wheel and the axle. Ans. As 12 is to 1. 

7. The angle of a screw is 30°, and the length of the power-arm 
is n times the radius of the cylinder. Find the mechanical advan- 
tage. Ans. n-j/"^ 

Section 186. 

1. If I run 100 yards in 11 seconds, what is my velocity in vels ? 

Ans. 27 3-11 vels. 

2. A man runs at the rate of 7 vels ; how long would he take to 
run a mile with this velocity ? Ans. 12 min. 34 2-7 sees. 

3. Express in miles per hour (1) 40 vels, (2) 100 yards per min- 
ute, a |(1) 27 3-11 miles per hour. 

( (2) 3 9-22 miles per hour. 

4. How many vels has the extremity (1) of the minute hand of 

a clock which is 1 foot long, (2) of the hour hand which is 10 inches 

22 
long. [The circumference of a circle = —z-oi the diameter.] 

Ans. (1) vels. (2) — — vels. 

7 6300 v ' 90720 

5. How many vels has a man standing at the equator, in conse- 
quence of the daily revolution of the earth about its axis, the diam- 
eter of the earth being, say, 8000 miles ? . , 32 . 

& ' J ' Ans. 1536 — vels. 

63 

6. The velocity of sound is 1120 vels ; how long does it take to 

travel 13 miles ? . , . , ] 7 

Ans. 1 min. 1 — sees. 
43 



n 

7. Which is the greater velocity 60 miles an hour or 500 yards 
in 11^ seconds ? 

Ans. 500 yards in 11^ see. ; their ratio is 66 to 100. 

8. Find the ratio of the velocities 2^ miles in 3 minutes and 10 
feet in a quarter of a second. Ans. 11 to 6. 

9. The following are the records in foot-racing : (1) 100 yards 
in 10 seconds ; (2) a quarter of a mile in 49 seconds ; (3) a mile in 
4 minutes 12 3-5 seconds ; (4) ten miles in 51 minutes 6 3-5 seconds ; 
express the average velocity of each in vels. Ans. (1) 30 vels ; 

20 — miles per hour ; (2) 26 -— vels ; 18 — miles per hour. (3) 
11 F ' v ; 49 ' 49 F v ; 

380 . , 106 .. , ..1113 . ,3779 .. 

20 vels: 14 — miles per hour. (4) 17 — - vels: 11 miles 

421 421 r w 5111 5111 

per hour. 

10. The following are the records in bicycle racing : (1) A mile 
in 2 minutes 30 seconds ; (2) 20 miles in 59 minutes 6 3-5 seconds ; 
(3) 22 miles 150 yards in 1 hour ; express the average velocity of 
each in vels and in miles per hour. Ans. (1) 35 1-5 vels ; 24 

miles per hour ; (2) 29 vels ; 20 — - miles per hour ; (3) 32 

r 5911 ' 5911 F 360 

15 
vels ; 22 — miles per hour. 
176 ' 

Section 187. 

1. Express in vels a velocity of c miles per hour. 

a 22 c . 

Ans. - — vels. 

15 



2. A point has v vels. How far does it go in h hours ? 



15 hv 

Ans. miles. 

22 



3. A point has k vels. How long does it take it to go m yards ? 



a 3m 1 

Ans. -r— seconds, 
k 



4. A point goes m yards in t seconds. How many miles does it 
go in h hours, supposing its velocity uniform ? 



45 hm ., 

Ans. miles. 

22 t 



18 

5. A point goes X inches in 1 minute. How far does it go in k 
days at the same rate ? Ans. 40 I k yards. 

6. A point goes k feet in t seconds. How long does it take to 

go m miles with the same velocity ? . 22 mt , 

& J Ans. — hours. 

15 k 

7. A train goes n miles in h hours. How far does it go in t 

seconds? 22 nt 

Ans. — yards. 

45 k J 

Section 192. 

1. A point P. starts along a line with constant acceleration a 
eels, and t seconds later a point Q starts along the same line with 
constant velocity v vels ; prove that Q will not overtake P unless v 
is greater than a t. 

2. A point moves from rest over a certain distance with 32 eels ; 

it describes — of the whole distance during the last second of the 
25 ? 

motion ; find the whole interval occupied. Ans. 2j seconds. 

3. Prove that the distances passed over in successive equal inter- 
vals by a point starting from rest and moving with a constant acce- 
leration are proportional to the series 1, 3, 5, 7. 

4. An engine-driver, whose train is traveling at the rate of 30 
miles an hour, sees a danger signal at the distance of 220 yards, 
and does his best to stop the train ; supposing that he can stop the 
train when traveling 30 miles an hour in 440 yards, show that his 
train will reach the danger signal with a velocity of 21 1-5 miles 
per hour nearly. 

5. A stone is let fall from the top of a tower 256 feet high. In 
how many seconds will it reach the ground ? Ans. 4 seconds. 

Section 209. 

1. How long will it take a body to descend 100 feet on a plane 
whose length is 150 feet and whose height is 60 feet ? 

2. There is an inclined railroad track, 2^ miles long, whose in- 
clination is 1 in 35. What velocity will a car acquire, in running 
the whole length of the road by its own weight ? 

3. A body weighing 5 pounds descends vertically, and draws a 
weight of 6 pounds up a plane whose inclination is 45°, how far 
will the first body descend in 10 seconds ? 



19 

Section 210. 

1. What is the time of flight of a projectile in a vacuum when 
the angle of projection is 45° and the range 6000 feet ? 

Ans. 19.3 sees. 

2. What is the range of a projectile when the angle of projec- 
tion is 30° and the initial velocity 200 feet ? Ans. 10*76. 

3. The angle of projection under which a shell is thrown is 32° 
and the range 3250 feet. What is the time of flight ? 

Ans. 11.25 sees. 

4. Find the angle of projection and velocity of projection of a 
shell so that its trajectory shall pass through two points, the co- 
ordinates of the first being x == 1700 feet, y= 10 feet, and of the 
second, x = 1800 feet, y = 10 feet. 

Ans. a = 39 / 19 // ; v = 2218.3. 

5. At what elevation must a shell be projected with a velocity 
of 400 feet that it may range 7500 feet on a plane which descends 
at an angle of 30° ? Ans. a = 4° 34' 10" ; a = 55° 25' 41". 

Section 266. 

1. What is the weight of a body whose mass is 20 pounds? 

Ans. Wr=644. 

2. What is the mass of a body whose weight is 20 poundals. 

Ans. M = 62 pounds. 

3. Compare two masses at the same place on the earth's sur- 

W W WW 

face. M,= — 1 ;M 2 = — 2 ; M, : M 2 : : — l : — \ hutg =g 2 . • . M oo 

9i 9* 9, 9, 

W, which was proved experimentally by Newton. 

4. If we take g = 1 M W. 

W W 

5. If gravity is not constant M • M • : — 1 : — 2 . 

9 9 

6. What force produces in 1 gramme mass freely falling an 
acceleration of 981 centimetres per second per second! 

F = mg ; F = 1 X 981 = 981 dynes. 

7. A poundal as the unit of force is equal to the weight of what 
part of a pound mass ? 

F — mg ; F = 1 X 32.2 poundal ; 1 poundal = — — of pound 

mass = about half a ounce. 



20 

Section 269. 

1. What will a body weigh if carried 5 miles above the earth's 
surface ? 

2. Where is the weight of a body the greatest ? 

3. How is the weight of a body affected by carrying it over 
the surface of the earth ? 

Section 272. 

1. What would a 100 pounds body w r eigh half way down to the 
centre of the earth ? 

2. What would it weigh at the centre ? 

3. Beneath the earth's surface, how long would it take a body 
to fall the first 100 miles downward from the surface? What 
velocity would it acquire ? How long would it take to fall to the 
centre, and what velocity would it acquire ? 

4. In the last example, how would the times and velocities dif- 
fer if the force of gravity at the surface were to remain constant ? 

Section 273. 

1. How much more will a body weigh on the sun than on the 
earth? Ans. 27.9. 

2. How far would a body fall in one second on the sun. 

Section 274. 

1. A ball weighing 10 pounds is whirled around in a circumfer- 
ence of 10 feet radius, with a velocity of 30 feet per second, what 
is the tension upon the cord which restrains the ball ? 

2. With what velocity must a body revolve in a circumference 
of 5 feet radius, in order that the centrifugal force may equal the 
weight of the body ? 

3. A ball weighing 2 pounds is whirled round by a sling 3 feet 
long, making 4 revolutions per second. What is its centrifugal 
force ? 

4. A weight of 5 pounds is attached to the end of a cord 3 feet 
long and is just capable of sustaining a weight of 100 pounds. How 
many revolutions per second must the body make in order that the 
cord may be upon the point of breaking ? 



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